Amazon .NET Developer Interview Questions & Answers

I would build prefix sums for both arrays so I can calculate the two sums of any contiguous segment in constant time. Then I try segment lengths from n down to 1. For each length, I test every possible contiguous segment and calculate totalPowerOutput = sumPower + sumBoost * length. The first segment within powerMax gives the maximum possible length, so I return immediately. The worst-case time is O(n²), and the auxiliary space is O(n).

We have two arrays that describe the same computers. For any adjacent group, we add its normal power values. We also add its boosted values and multiply that boosted sum by the number of computers in the group. The group is valid when this total is at most powerMax. We need the largest valid adjacent group. I use prefix sums so each group's two sums are quick to calculate, then I test group sizes from largest to smallest. The first valid size is the answer.

1. Can powerOutputs or boostedOutputs contain negative values?
2. Can the arrays be empty?
3. Are the input values and powerMax small enough for 64-bit arithmetic, or should overflow beyond long be considered?

The inputs are powerOutputs, boostedOutputs, and powerMax. Matching positions in the two arrays describe the same computer. We must return a count, not indices or values. The count is the maximum length of a contiguous segment whose total power is at most powerMax.

For a segment of length L, the calculation used by the approved diagram is: totalPowerOutput = sumPower + sumBoost * L.

I build one prefix array for powerOutputs and one for boostedOutputs. A prefix sum stores the running total before each position. This lets me calculate the sum of any window [left, right) in O(1) time.

I then try lengths n, n - 1, and so on down to 1. For each length, I test every contiguous window. The main invariant is that before I move to a smaller length, every possible window of every larger length has already failed. Therefore, the first valid window proves that its length is the maximum.

This method does not require the array values to be nonnegative.

For the diagram example: powerOutputs = [2, 1, 4, 3, 2] boostedOutputs = [1, 3, 1, 2, 1] powerMax = 16

The prefix arrays are: prefixPower = [0, 2, 3, 7, 10, 12] prefixBoost = [0, 1, 4, 5, 7, 8]

For a window [left, right), I calculate: sumPower = prefixPower[right] - prefixPower[left] sumBoost = prefixBoost[right] - prefixBoost[left].

First I test length 5. Window [0..4] has sumPower = 12 and sumBoost = 8. Its total is 12 + 8 * 5 = 52, which is greater than 16, so it fails.

Next I test length 4. Window [0..3] gives 10 + 7 * 4 = 38. Window [1..4] also gives 10 + 7 * 4 = 38. Both fail.

Next I test every length-3 window. Window [0..2] gives 7 + 5 * 3 = 22. Window [1..3] gives 8 + 6 * 3 = 26. Window [2..4] gives 9 + 4 * 3 = 21. All are greater than 16.

Then I test length 2. Window [0..1] has sumPower = 3 and sumBoost = 4. Its total is 3 + 4 * 2 = 11. Since 11 <= 16, this window is valid. I return 2 immediately. No later length-2 windows or length-1 windows are processed.

Lengths are tested from largest to smallest. For each length, every possible contiguous window is checked before moving to the next smaller length. By the time the algorithm reaches length 2, lengths 5, 4, and 3 have all been completely ruled out. The first valid length-2 window therefore proves that no longer valid segment exists. The prefix arrays also calculate each tested window's normal-power and boosted-power sums exactly.

The code first validates null inputs and requires the arrays to have the same length. It builds two long prefix arrays. It then uses an outer loop for lengths from n down to 1 and an inner loop for each possible left boundary. right is left + length, so the code represents the current segment as the half-open range [left, right). It gets both sums from the prefix arrays, calculates totalPowerOutput, and returns the length as soon as that value is at most powerMax. If no segment works, it returns 0.

Building the prefix arrays takes O(n) time. The nested search tests O(n²) windows in the worst case, and each test takes O(1) after the prefix arrays are built. Therefore, the worst-case time is O(n²). The two prefix arrays require O(n) auxiliary space. Relevant edge cases include empty arrays, one computer, negative values, unequal array lengths, and large totals where 64-bit arithmetic should be considered carefully.

The key insight is to make each contiguous-window sum cheap and then search candidate lengths in an order that makes early return correct. Two prefix-sum arrays let us calculate sumPower and sumBoost for any window [left, right) in O(1). We test candidate lengths from n down to 1 and examine every possible window of each length. The invariant is that before a smaller length is considered, all windows of every larger length have already been proven invalid. Therefore, the first feasible window has the maximum possible length. This approach also remains correct when values are negative because it does not depend on sliding-window monotonicity.

Let n be the number of computers. Building the two prefix arrays takes O(n) time. After that, each window's normal-power sum and boosted-power sum take O(1) time to calculate. In the worst case, the nested loops test O(n²) windows, so the total worst-case time is O(n²). The prefixPower and prefixBoost arrays each grow with n, so the auxiliary space is O(n). The algorithm can finish earlier if it finds a valid window before all candidate lengths are tested.

One common mistake is using a normal sliding window without a guarantee that the values make the cost monotonic. The approved solution avoids that assumption. Another mistake is applying the formula incorrectly by multiplying only one boosted value instead of the boosted sum for the whole segment. Candidates can also mix inclusive example indices such as [0..1] with the code's half-open [left, right) prefix-sum boundaries. Another mistake is returning a valid segment before larger lengths have been ruled out. Finally, candidates may claim O(n) time even though the approved nested search is O(n²) in the worst case.

State the invariant before coding: "I test lengths from largest to smallest, and I check every window of a length before moving down, so the first valid window must have the maximum possible length." Then explain that prefix sums make each individual window calculation O(1).

I would sort the integers first. Then I keep the largest N minus 1 values as singleton arrays, so each of those values is guaranteed to be its array's median. I put every remaining value into one final array and use that array's median. Then I add the N minus 1 singleton medians. This keeps the smaller values from lowering several large medians. Sorting takes O(M log M) time. The algorithm uses O(1) auxiliary space beyond the sort implementation when only the maximum sum is returned.

We need to split all integers into exactly N non-empty arrays. Every input value must belong to one array. We want the sum of the N array medians to be as large as possible. The diagram uses a sorting and greedy approach. We protect the largest N minus 1 values by making each one a single-element array. All remaining values go into one final array. This means the smaller values can affect only one median instead of lowering several large medians. The diagram assumes that an even-length array uses the average of its two middle sorted values.

1. How should the median be defined when an array has an even number of elements?
2. Should I return only the maximum sum, or also the actual N arrays?
3. Is it acceptable to sort the input array in place?

The input is an integer array called nums and an integer N. We must use every integer exactly once and create exactly N non-empty arrays. The required result is the maximum possible sum of the N medians. The approved diagram assumes that for an even-sized array, the median is the average of the two middle sorted values.

Let M be nums.Length. Sort nums in non-decreasing order. Keep the largest N minus 1 values as N minus 1 singleton arrays. A singleton array contains one value, so that value is automatically its median. Put every other value into one final array. These remaining values form a sorted prefix of nums, so we can read that final array's median directly.

The input is nums = [12, 1, 5, 10, 3, 8, 2, 9, 6, 11, 4, 7] and N = 4. After sorting, nums is [1,2,3,4,5,6,7,8,9,10,11,12]. Here M = 12. We need N minus 1 = 3 singleton arrays. The split index is M - (N - 1) = 12 - 3 = 9. Sorted indices 9, 10, and 11 contain 10, 11, and 12. They become singleton arrays. Sorted indices 0 through 8 form [1,2,3,4,5,6,7,8,9]. Its length is 9, so its zero-based middle index is 4 and its median is 5. The four medians are 10, 11, 12, and 5. Their sum is 10 + 11 + 12 + 5 = 38.

The key invariant is that the largest N minus 1 values remain singleton medians. Each of those large values is therefore guaranteed to contribute its full value to the result. All smaller values are grouped into the one remaining array, so only that array's median is affected by them. Placing smaller values into one of the large singleton arrays cannot increase that singleton's median and can lower it. The greedy construction therefore protects the largest median contributions while concentrating the effect of the smaller values into one array.

The method first validates nums and N. It sorts nums in ascending order. remainingCount = M - (N - 1) is the number of values in the final non-singleton array. Because that final array is the sorted prefix, the code reads its median directly. For an odd remainingCount, it reads one middle value. For an even remainingCount, it averages the two middle values according to the diagram's stated assumption. It then adds every value from remainingCount through M - 1. Those values are the N minus 1 singleton medians. Finally, it returns the total.

Sorting M values takes O(M log M) time. Reading the remaining-array median is O(1), and adding the N minus 1 singleton medians takes O(N), so sorting dominates. The algorithm uses O(1) auxiliary space beyond the sort implementation when only the maximum sum is returned. If the actual arrays are materialized, storing them requires O(M) additional space. If N = 1, the result is the median of the whole sorted input. If N = M, every array is a singleton and the result is the sum of all values. Duplicate and negative integers follow the same greedy structure. N less than 1 or greater than M is invalid.

Sort the integers in non-decreasing order. Let M be the number of integers. Reserve the largest N minus 1 values as N minus 1 singleton arrays. Put the first M - (N - 1) sorted values into the final array. The central invariant is that the largest N minus 1 values stay protected as singleton medians, while every smaller value affects only the one remaining array. Because the final array is already sorted, its median can be read directly from the sorted prefix. Add that median to the N minus 1 singleton values to get the maximum sum shown by the diagram.

Let M = nums.Length. Sorting the M integers takes O(M log M) time. After sorting, finding the final-array median takes O(1) time. Adding the N minus 1 singleton medians takes O(N) time, so the sorting step is the dominant cost. The algorithm uses O(1) auxiliary space beyond the sort implementation when it returns only the maximum sum. If we also build and return the actual partition arrays, storing those arrays requires O(M) additional space.

One mistake is choosing all N largest values as medians and then placing smaller values into those arrays. Adding smaller values can change a median, so those large values may no longer remain medians. Another mistake is using the wrong split index. The final array must contain M - (N - 1) values. Candidates may also forget to clarify how an even-length median is calculated. Another common error is forgetting the O(M log M) sorting cost. Finally, do not claim O(1) auxiliary space if the implementation actually materializes and stores all N partition arrays.

State the invariant before coding: the largest N minus 1 values stay as singleton medians, and every remaining value goes into one final array. Then derive remainingCount = M - (N - 1). This connects the greedy proof directly to the implementation.

I would reverse the tree with recursion. At each non-null node, I swap its left and right child references. Then I recursively process the new left subtree and the new right subtree. This gives the pre-order sequence 1, 3, 7, 6, 2, 5, 4 for the shown example. Every node is processed once, so the time complexity is O(n). The recursion stack uses O(h) auxiliary space, where h is the tree height.

The problem gives a binary tree and asks us to mirror it. Mirroring means every left child becomes the right child, and every right child becomes the left child. We keep the same root node. We only change the child links below it. The diagram uses recursion. At each node, we swap its two children first. Then we process the new left side and the new right side. This works because the same rule is applied to every node in the tree.

1. Can the input root be null?
2. Is it acceptable to modify the existing tree instead of creating a new tree?
3. Should I return the same root reference after the tree is mirrored?

The input is the root node of a binary tree. The output is the same root node after the whole tree has been mirrored. We are changing node references, not node values. The root stays the same object. For the example, the original tree has root 1. Node 1 has children 2 and 3. Node 2 has children 4 and 5. Node 3 has children 6 and 7. The mirrored result has root 1, children 3 and 2, then children 7 and 6 under node 3, and 5 and 4 under node 2.

For each node, first check whether it is null. If it is null, return null. Otherwise, swap the left and right child references. After the swap, recursively reverse the new left subtree. Then recursively reverse the new right subtree. Finally, return the current node. The important rule is that when a node finishes, its own children have been swapped and both of its subtrees have also been mirrored.

We start at node 1. Its children are 2 and 3. We swap them, so node 1 now has left child 3 and right child 2. Because the code recurses into the new left child first, the next node is 3.

At node 3, its children are 6 and 7. We swap them, so its children become 7 and 6. We then visit node 7. It is a leaf, so swapping its null children does not change the tree. Its recursive calls reach null and return. Next we visit node 6. It is also a leaf and returns without a structural change.

After the subtree rooted at 3 is complete, we process node 2. Its children are still 4 and 5 at this point. We swap them, so they become 5 and 4. Then node 5 is processed as a leaf, followed by node 4 as a leaf. The complete processing order is 1, 3, 7, 6, 2, 5, 4.

The final tree is root 1 with left child 3 and right child 2. Node 3 has children 7 and 6. Node 2 has children 5 and 4. This matches the mirrored tree in the diagram.

At every non-null node, the algorithm swaps exactly the two child references that must exchange sides in the mirrored tree. Then it applies the same operation to both resulting subtrees. The base case stops recursion when there is no node. Because every node is handled this way, every left and right relationship in the original tree is reversed. The root reference itself never changes.

The method receives a TreeNode reference. If the reference is null, it returns null. Otherwise, it saves the original left child in a temporary variable, assigns the original right child to Left, and assigns the saved child to Right. It then calls ReverseTree on root.Left and root.Right in that order. Those calls mirror both subtrees. Finally, the method returns root, which is the same node reference that was passed into that recursive call.

There are n nodes in the tree. Each node is visited and processed once, so the time complexity is O(n). The recursion stack can contain one call for each level of the tree, so the auxiliary space is O(h), where h is the tree height. In the worst case of a completely skewed tree, h can be n, giving O(n) stack space. A null tree returns null. A single-node tree stays unchanged. A skewed tree is still mirrored correctly. A complete or any arbitrary binary tree is handled by the same recursive rule.

The key idea is to mirror the tree locally at every node. For the current node, swap its left and right child references. Then recursively apply the same operation to the new left subtree and the new right subtree. The central invariant is: when ReverseTree returns for a non-null node, that same node is still the subtree root, its left and right children have exchanged sides, and both child subtrees are already mirrored. The null base case stops recursion at the ends of the tree.

Let n be the number of nodes and h be the height of the tree. The algorithm visits and processes every node once, so the time complexity is O(n). The extra memory comes from recursive calls. At most one chain from the root to a deepest node is active at one time, so the auxiliary space is O(h). For a balanced tree, h is about log n. For a completely skewed tree, h can be n, so the worst-case recursion space is O(n).

A common mistake is to forget the null base case, which can cause recursion to fail when it reaches missing children. Another mistake is to swap the children but then recurse using the wrong assumed order. In this solution, the swap happens first, so the recursive order follows the new left child and then the new right child. Candidates can also accidentally create a new root even though the diagram keeps the same root reference. Another mistake is to say the auxiliary space is O(1) and ignore the recursion stack. Finally, changing node values instead of swapping child references does not correctly mirror the tree structure.

Say the order out loud while coding: "swap first, recurse into the new left child, recurse into the new right child, return the same root." This makes it easier to keep your code and walkthrough consistent.

I would use expand around center. Every palindrome has a center, so for each index I check an odd-length center at (i, i) and an even-length center at (i, i + 1). I expand left and right while the characters match. When I find a longer palindrome, I save its start index and length. For "babad", the method returns "bab", while "aba" is also valid. The worst-case time is O(n²), and the auxiliary space is O(1).

We are given one string and need to return its longest contiguous part that reads the same forward and backward. That part is called a palindrome. A palindrome can have one character in the middle, such as "bab", or a center between two characters, such as "abba". The diagram uses expand around center because every palindrome has one of these two center forms. We test both forms at every position and remember the longest palindrome found.

1. If more than one longest palindrome exists, can I return any one of them?
2. Should an empty input return an empty string?
3. Should character comparison remain case-sensitive?

The input is a string s. We must return one contiguous substring of s that is a palindrome and has maximum length. We return the substring itself, not its indexes. For s = "babad", both "bab" and "aba" have length 3. The shown implementation keeps the first longest palindrome it finds, so its returned result is "bab".

Every palindrome has a center. An odd-length palindrome has one character as its center. An even-length palindrome has a center between two characters. For each index i, the algorithm first checks the odd center (i, i). It then checks the even center (i, i + 1). From each center, it moves left and right while both indexes remain inside the string and the two characters are equal.

If the string is null, empty, or has one character, the method returns it immediately. Otherwise, n is the string length. bestStart starts at 0 and bestLen starts at 1. A single character is always a palindrome, so this is a valid initial answer. For each center, left and right mark the current palindrome boundaries.

For s = "babad", the characters are b, a, b, a, d at indexes 0 through 4.

At odd center i = 0, s[0] equals s[0], so "b" is a palindrome of length 1. It does not beat bestLen = 1. After moving outward, left becomes -1, so expansion stops.

At even center (0, 1), s[0] is "b" and s[1] is "a". They do not match, so no even-length palindrome is found there.

At odd center i = 1, s[1] equals s[1], giving "a". The next comparison is s[0] with s[2]. Both are "b", so the palindrome becomes "bab" with length 3. After moving outward again, left becomes -1. Because 3 is greater than bestLen = 1, the saved state becomes bestStart = 0 and bestLen = 3.

At even center (1, 2), s[1] and s[2] are different, so expansion stops immediately.

At odd center i = 2, s[2] equals s[2], giving "b". Next, s[1] equals s[3], so the palindrome becomes "aba" with length 3. Next, s[0] is "b" and s[4] is "d", so expansion stops. Length 3 only ties the saved best length. The code updates only when currentLen is strictly greater than bestLen, so "bab" remains saved.

The remaining centers do not produce a longer palindrome. At odd center i = 4, s[4] equals s[4], giving "d". After moving outward, right becomes 5, which is outside the string, so expansion stops. The final state is bestStart = 0 and bestLen = 3. Therefore, s.Substring(0, 3) returns "bab".

Every palindrome is either odd-length or even-length, so every palindrome has one of the center forms that the algorithm checks. For one chosen center, expanding while the two characters match finds the largest palindrome for that center. Since the algorithm examines every possible center and keeps the longest palindrome seen, the saved substring is a longest palindromic substring.

The main loop visits every index from 0 through n - 1. For each i, it calls ExpandAroundCenter twice. The first call uses left = i and right = i for an odd-length center. The second uses left = i and right = i + 1 for an even-length center. The helper checks the bounds and character equality before each expansion. It calculates currentLen = right - left + 1 and updates bestStart and bestLen only when currentLen is larger. Finally, the method returns s.Substring(bestStart, bestLen).

There are O(n) possible centers. In the worst case, expansion from one center can inspect O(n) characters. Therefore, the total worst-case time is O(n²). The algorithm uses only a constant number of integer variables, so auxiliary space is O(1). Important cases shown in the diagram are an empty string, one character, all identical characters, and a string with no repeated characters.

The key idea is that every palindrome has a center. An odd-length palindrome is centered on one character, while an even-length palindrome is centered between two characters. For each index, we expand from both center types. The important invariant is that the range from left through right is a palindrome whenever the indexes are valid and s[left] equals s[right]. Expanding until a mismatch or boundary finds the longest palindrome for that center. Checking all centers and keeping the longest result therefore finds a longest palindromic substring.

Let n be the string length. There are O(n) centers to examine. In the worst case, one center can expand across O(n) characters. This gives O(n²) worst-case time. The algorithm does not use a map, set, stack, queue, dynamic programming table, or copied working array. It keeps only a constant number of indexes and lengths, so the auxiliary space complexity is O(1).

A common mistake is checking only odd-length centers, which misses even-length palindromes such as "abba". Another is checking only even-length centers and missing values such as "bab". Candidates may confuse a substring with a subsequence, but this problem requires contiguous characters. It is also easy to calculate the inclusive length incorrectly; it must be right - left + 1. Another mistake is reading s[left] or s[right] before checking the bounds. Finally, changing the saved answer on an equal-length tie would not match the shown strict greater-than update rule.

Before coding, draw one odd center and one even center. Then explain that the helper expands outward while characters match, saves only a strictly longer palindrome, and stops at a mismatch or string boundary.

I would scan the grid row by row. When I find a land cell that has not been visited, I increase the island count and run DFS from that cell. DFS marks every horizontally or vertically connected land cell as visited, so the same island is never counted twice. Diagonal cells are not connected. Each cell is processed at most once, so the time complexity is O(M × N). The visited matrix and recursion stack use O(M × N) auxiliary space.

The input is a grid with M rows and N columns. Each cell is either land or water. We need to count how many separate groups of land exist. Land cells belong to the same island only when they touch up, down, left, or right. Diagonal touching does not connect islands. I scan the grid from top-left to bottom-right. Whenever I find land that has not been visited, I count one new island and use DFS to mark every connected land cell in that island as visited.

1. Should diagonal land cells be considered connected? In this solution, diagonal adjacency is excluded.
2. Can the grid be empty? The implementation handles an empty grid by returning 0.

The input is a two-dimensional grid. A land cell is represented by '1' and a water cell by '0'. We return one integer: the total number of islands. Two land cells belong to the same island only when they are connected through up, down, left, or right. Diagonal neighbors do not connect islands.

I use depth-first search, or DFS. DFS starts from one unvisited land cell and reaches every land cell in the same four-directionally connected component. I also use a bool[,] visited matrix with the same M by N size as the grid. visited[r, c] tells us whether that cell has already been processed. The main invariant is that after one DFS finishes, every land cell in that island is marked visited.

Set islandCount to 0. Create the visited matrix with every value initially false. Scan the cells in row-major order, from the top-left toward the bottom-right. For each cell, check whether it is land and has not been visited. If both conditions are true, increment islandCount and start DFS. DFS finishes marking the complete connected island before the outer scan continues.

The example grid is [[1,1,0,0,1],[1,0,0,1,1],[0,0,0,0,0],[1,0,1,1,0]]. It has 4 rows, 5 columns, and 20 cells. The first unvisited land cell is (0,0). DFS visits (0,0), (0,1), and (1,0), so islandCount becomes

1. The next unvisited land cell is (0,4). DFS visits (0,4), (1,4), and (1,3), so islandCount becomes
2. The next unvisited land cell is (3,0). It is isolated, so islandCount becomes
3. The next unvisited land cell is (3,2). DFS also visits (3,3), so islandCount becomes
4. No unvisited land remains, so the returned result is 4.

Every DFS begins from one land cell that has not been visited. That DFS marks every land cell reachable from it using only the four allowed directions. Because all cells in that island become visited, the outer scan cannot count the same island again. Every island contains at least one land cell that the scan eventually reaches. Therefore each island starts exactly one DFS and increases islandCount exactly once.

The Main method creates the exact 4 by 5 example grid and prints the result. NumIslands handles an empty grid, creates the visited matrix, defines the four allowed direction offsets, and scans every row and column. When it finds unvisited land, it increments count and calls Dfs. Dfs stops for coordinates outside the grid, water cells, or already visited cells. Otherwise it marks the current cell visited and recursively explores up, down, left, and right.

There are M × N cells. Each cell is processed at most once as part of the traversal, so the time complexity is O(M × N). The visited matrix requires O(M × N) space. In the worst case, such as one large all-land island, the recursive DFS call stack can also grow to O(M × N). Relevant edge cases include an empty grid, an all-water grid, an all-land grid, a single land cell, and land cells that touch only diagonally.

Use DFS to count four-directionally connected components. Scan the grid in row-major order. When grid[r][c] is land and visited[r, c] is false, that cell starts a new island. Increment the island count and run DFS from that cell. DFS marks the current cell and every land cell reachable through up, down, left, or right as visited. The central invariant is that after a DFS finishes, every cell in that island has been marked visited. Therefore the same island cannot be counted again.

Let M be the number of rows and N be the number of columns. The time complexity is O(M × N) because there are M × N cells, and each cell is processed at most once by the traversal. The auxiliary space is O(M × N). The visited matrix contains M × N Boolean entries. In the worst case, the recursive DFS call stack can also grow to O(M × N), such as when the grid contains one very large connected island.

A common mistake is treating diagonal land cells as connected even though diagonal adjacency is excluded. Another mistake is incrementing the island count for every land cell instead of only when an unvisited land cell starts a new DFS. Marking a cell visited too late can cause repeated recursive work or cycles between neighboring land cells. Candidates may also forget the out-of-bounds, water, or already-visited base cases. Finally, claiming O(1) auxiliary space is incorrect because the visited matrix and recursive call stack can grow with the grid.

State the invariant before coding: every time DFS starts from an unvisited land cell, you have found exactly one new island, and that DFS marks the whole island so it cannot be counted again.

I would use breadth-first search, or BFS, from the first E cell. I keep a queue, mark each cell visited when I enqueue it, and store its parent so I can rebuild the route later. BFS explores the grid by increasing number of moves, so the first time I dequeue the second E, I have a shortest path. Blocked X cells are skipped. The time complexity is O(N × M), and the auxiliary space is O(N × M).

The matrix contains two E cells. One is the start and one is the goal. A cell marked _ can be used. A cell marked X cannot be used. We may move only up, down, left, or right. We need the route with the smallest number of moves. The diagram uses breadth-first search because every valid move has the same cost. It also stores parent links so the actual shortest route can be returned.

1. Should I return the actual path as well as its length?
2. Can I assume, as stated, that there are exactly two E cells and that a valid path always exists?
3. If several shortest paths exist, is returning any one of them acceptable?

The input is an N by M character matrix. Each cell is E, _, or X. There are exactly two E cells. In the diagram, the first E found in row-major order is the start at (0,0). The second E is the goal at (5,5). We return one shortest route and its length in moves. For the shown 6 by 6 example, the result is 10 moves and 11 cells.

I use breadth-first search. BFS visits reachable cells in increasing distance from the start. A queue stores cells waiting to be processed. I also use a visited matrix so a cell is added only once. A parent matrix stores which cell led to each newly discovered cell. The main invariant is that when a cell is first discovered by BFS, the recorded route to that cell has the minimum possible number of moves.

First, scan the matrix to find the two E cells. Create a queue, a visited matrix, and a parent matrix. Put the start cell into the queue and mark it visited immediately. The parent of the start remains null because it has no previous cell. The implementation checks the four legal moves in the order up, down, left, and right. It rejects cells that are outside the matrix, contain X, or were already visited.

The start is (0,0). The distance levels shown in the diagram are: level 0: (0,0). Level 1: (0,1). Level 2: (0,2). Level 3: (1,2). Level 4: (2,2). Level 5 contains (2,3) and (2,1). Level 6 contains (2,4) and (2,0). Level 7 contains (3,4), (1,4), (2,5), and (3,0). Level 8 contains (4,4), (0,4), and (4,0). Level 9 contains (4,5), (4,3), (0,5), and (4,1). At distance 10, the goal (5,5) is reached. The listed cells describe BFS distance levels. Cells on the same level all have the same minimum distance.

One shortest route shown in the diagram is (0,0) → (0,1) → (0,2) → (1,2) → (2,2) → (2,3) → (2,4) → (3,4) → (4,4) → (4,5) → (5,5). This route has 11 cells, so it has 10 moves.

Every legal move has the same cost of one step. BFS processes cells by increasing distance from the start. Therefore, when a cell is first discovered, no shorter route to that cell can appear later. When the goal is dequeued, its stored parent chain represents a shortest route. We then follow parent links from the goal back to the start and reverse that list.

The code scans for the two E cells. It allocates bool[,] for visited and Cell?[,] for parent links. It enqueues the start and marks it visited before processing begins. For each dequeued cell, it checks up, down, left, and right. A valid unvisited neighbor is marked visited, given a parent, and enqueued. Processing stops when the goal is dequeued. Finally, the code follows parent links backward, reverses the collected cells, and returns path.Count - 1 as the number of moves.

There are N × M cells. Each cell is enqueued at most once, and each processed cell checks four directions. Therefore the time complexity is O(N × M). The visited matrix, parent matrix, queue, and reconstructed path can all grow with the matrix, so auxiliary space is O(N × M). Relevant cases include a single-row or single-column route, adjacent E cells, several equally short paths, large open areas with narrow corridors, and E cells that are not on the border.

Treat each traversable grid cell as a node in an unweighted graph. A move to an adjacent non-X cell is an edge with cost one. BFS is appropriate because all allowed moves have equal cost. The queue processes cells by increasing shortest distance from the start. The central invariant is that when BFS first discovers a cell, its parent chain represents a shortest route to that cell. Marking a cell visited when it is enqueued prevents duplicate queue entries. Parent links allow one shortest route to be reconstructed after reaching the goal.

Let N be the number of rows and M be the number of columns. There are N × M cells. Each cell can enter the queue at most once, and every processed cell checks only four directions. The time complexity is O(N × M). The visited matrix, parent matrix, BFS queue, and reconstructed path can together use O(N × M) extra memory. Auxiliary space means memory used by the algorithm in addition to the input matrix.

A common mistake is using DFS and assuming the first route found is the shortest. Another is marking a cell visited only when it is dequeued, which can put the same cell in the queue multiple times. Candidates may forget to reject X cells or accidentally allow diagonal moves. Another mistake is not storing parent links when the actual path must be returned. Finally, counting cells instead of edges gives 11 instead of the correct 10 moves in the diagram example.

State the BFS invariant before coding: the queue explores cells by increasing distance, and each cell is marked visited when it is enqueued. Then explain that the parent matrix is used only to rebuild one shortest route. This makes the correctness argument and code structure easy to follow.

I would compare the strings from right to left using two pointers and two skip counters. When I see '#', I increase that string's skip count. While the skip count is positive, I ignore earlier characters because they are deleted. I only compare characters that survive the backspaces. If the current valid characters differ, I return false. If both strings finish together, I return true. This takes O(n + m) time and O(1) auxiliary space.

We have two strings where '#' acts like a backspace. It removes the previous character that has not already been removed. We need to decide whether both strings contain the same characters after all backspaces are applied. Instead of building two new strings, we start from the end of each input. We count pending backspaces and skip characters that would be deleted. Then we compare only the characters that remain. This fits the problem because deletions affect characters to the left, so scanning backward lets us process them directly.

1. Should '#' always mean one backspace that deletes the previous available character?
2. Can a string contain more backspaces than available characters?
3. Is character comparison case-sensitive?

The input is two strings, s and t. The output is a bool. We return true when the two strings are equal after applying every '#'. Otherwise, we return false. In the diagram, s is "ab#c" and t is "ad#c". Both become "ac", so the result is true.

I use pointer i for s and pointer j for t. Both start at the last character. I also use skipS and skipT. A skip counter means how many earlier characters must be ignored because of backspaces. The main invariant is that whenever the code reaches the comparison step, i and j either point to the next surviving character from the right or are already before the start of the string.

For s = "ab#c", i starts at 3. For t = "ad#c", j also starts at 3. Both skip counters start at 0. At index 3, both strings contain 'c'. These characters are valid, so we compare them. They match. Then both pointers move left.

At index 2, both strings contain '#'. We increase skipS and skipT from 0 to 1 and move both pointers left. At index 1, s contains 'b' and t contains 'd'. Because each skip counter is 1, those characters are deleted. We decrease both skip counters back to 0 and move left again. At index 0, both strings contain 'a'. They are valid and equal. After comparing them, both pointers move to -1. Both strings are exhausted, so the method returns true.

Each skip counter exactly represents how many earlier characters still need to be deleted. The algorithm never compares a deleted character. It compares surviving characters in the same right-to-left order. If two surviving characters differ, the processed strings cannot be equal. If both strings finish without a mismatch, every surviving character matched, so the processed strings are equal.

The outer loop continues while either string still has characters to process. Two inner loops find the next valid character in each string. A '#' increases the matching skip counter. A normal character is skipped when that counter is positive. After both inner loops finish, the code reads the current valid characters, using '\0' when a string is exhausted. If those characters differ, it returns false. Otherwise, both pointers move left. When the outer loop ends, it returns true.

If n is the length of s and m is the length of t, every character moves past a pointer at most once. The time complexity is O(n + m). The algorithm uses only two indices, two skip counters, and two temporary characters, so auxiliary space is O(1). It also handles empty strings, strings containing only backspaces, extra backspaces, different resulting lengths, long runs of '#', and strings with no backspaces.

The key idea is to avoid creating cleaned copies of the strings. Backspaces delete characters to their left, so scanning from right to left makes it possible to count deletions before deciding whether a character should be compared. Pointer i scans s and pointer j scans t. skipS and skipT record pending deletions. The central invariant is that after each inner skip loop finishes, the pointer either identifies the next surviving character from the right or is before the start of the string. Therefore, comparing those surviving characters is enough to determine equality. This keeps the extra memory constant.

Let n be s.Length and m be t.Length. The time complexity is O(n + m). Each pointer only moves from right to left, and each character is processed at most once. A character may be recognized as '#', skipped because of a pending backspace, or compared as a surviving character. The algorithm does not build new strings or collections. It uses two indices, two integer skip counters, and two character variables. Therefore, the auxiliary space complexity is O(1).

A common mistake is to compare characters before removing the effect of pending backspaces. Another is to move left after seeing '#' without remembering that an earlier character must also be skipped. Candidates may also forget that several '#' characters can create several pending deletions. Another mistake is assuming both processed strings have the same length and accessing a pointer that is already below zero. Finally, claiming O(n + m) extra space would be incorrect for this implementation because it does not build cleaned strings.

While coding, explain the invariant before writing the comparison: after each skip loop, the pointer is either on the next surviving character or before the beginning of the string. That makes the comparison logic much easier to justify.

I would build two hash-based indexes: each user to the set of products they bought, and each product to the set of users who bought it. For the target user, I find other users who share at least one purchased product. Then I collect products bought by those users and exclude products the target user already owns. The hash-based work is O(n) expected time. Because the diagram sorts the final recommendations, total time is O(n + r log r), with O(n) auxiliary space.

The input gives one target user and a list of purchases. Each purchase connects a user to a product. We need to recommend products the target user has not bought. A product is eligible only when another user bought it and that user shares at least one product with the target user. The diagram solves this by building two purchase indexes, finding overlapping users, collecting their products, and removing products already owned by the target user.

1. Should the returned products be unique? The diagram returns unique products.
2. Does the output need a stable order? The diagram sorts the result for deterministic output.
3. What should happen when the target user has no purchases? The diagram returns an empty list.

The target user is user 1. The purchases are (1, P1), (1, P2), (2, P2), (2, P3), (3, P1), (3, P4), and (4, P5). User 1 already owns P1 and P2. We must return products that user 1 does not own but that were bought by users who share at least one product with user 1.

First, build a user-to-products map. It contains 1 → {P1, P2}, 2 → {P2, P3}, 3 → {P1, P4}, and 4 → {P5}. Also build a product-to-users map. It contains P1 → {1, 3}, P2 → {1, 2}, P3 → {2}, P4 → {3}, and P5 → {4}. Hash sets also make duplicate purchase records harmless because each user-product relationship is stored once.

User 1 owns P1 and P2. From P1, the product-to-users map gives users {1, 3}. Excluding user 1 leaves user 3. From P2, it gives {1, 2}. Excluding user 1 leaves user 2. The candidate users are therefore {2, 3}. User 4 is not included because user 4 shares no product with user 1.

User 2 bought {P2, P3}. User 3 bought {P1, P4}. Their combined unique product set is {P1, P2, P3, P4}. User 1 already owns {P1, P2}, so those products are excluded. The remaining products are {P3, P4}.

P3 is valid because user 2 bought P3 and user 2 shares P2 with user 1. P4 is valid because user 3 bought P4 and user 3 shares P1 with user 1. Neither P3 nor P4 is already owned by user 1. The code sorts the final list, so the returned result is [P3, P4].

The C# code builds both dictionaries in one pass over the purchase list. It reads the target user's product set, discovers candidate users through the product-to-users map, gathers their products in another hash set, skips products already owned by the target user, converts the set to a list, and sorts it. Dictionary and HashSet operations are O(1) on average. The hash-processing work is O(n) expected time. Sorting r recommendations adds O(r log r), so total expected time is O(n + r log r). Auxiliary space is O(n).

The key idea is to represent the purchase history in both directions. The user-to-products dictionary tells us what each user owns. The product-to-users dictionary tells us who bought each product. The central invariant is that a candidate user is added only because that user bought at least one product already owned by the target user. Products are then collected only from those candidate users, and any product already in the target user's set is excluded. HashSet keeps candidate users and products unique. This directly enforces both rules in the question: a common purchase must exist, and an already-owned product must never be recommended.

Let n be the number of unique purchase relationships represented by the input after set deduplication, and let r be the number of recommended products. Building the two dictionaries and sets takes O(n) expected time because Dictionary and HashSet lookup and insertion are O(1) on average. Finding overlapping users and collecting their products also takes O(n) expected time over the stored relationships. The code then sorts the r returned products, which costs O(r log r). Therefore, the complete code runs in O(n + r log r) expected time. The dictionaries and sets use O(n) auxiliary space.

A common mistake is recommending products from every other user instead of only users who share at least one product with the target user. Another mistake is forgetting to exclude products the target user already owns. Using lists instead of sets can also produce duplicate users or duplicate recommended products. Another error is including the target user in the overlapping-user set even though the requirement refers to other users. Finally, claiming the complete code is O(n) ignores the final sort. The hash-processing part is O(n) expected time, while sorting adds O(r log r).

Explain the solution in two directions: user to products tells you what someone owns, and product to users tells you who overlaps with the target user. Then explain the two filters: first keep only users with a shared purchase, and then remove products the target user already owns. This makes the correctness argument easy to follow.

At a high level, I would complete the product API so every request follows one clear path. The client sends an HTTPS request through the reverse proxy. ASP.NET Core runs middleware, maps the route, and calls the correct controller. The controller calls ProductService, which uses ProductRepository and EF Core to access the SQL database. JWT authentication checks the caller, while authorization controls access. The result returns as JSON with the shown HTTP status codes. The benefit is clear separation of responsibilities. The trade-off is more layers and configuration to maintain.

This question asks me to finish a small product API and explain how all its parts work together. A client needs to read, create, update, or delete product information. Each request must reach the correct code. The application must process the request safely, access stored data, and send a clear answer back. The main challenge is keeping each responsibility in the right place. I will follow the diagram from the client, through the application, to the database, and then back to the client.

• Should I complete only the product endpoints already shown in the codebase?
• Should I keep the controller, service, repository, and EF Core structure shown here?
• Should authentication, logging, rate limiting, and error handling remain as shown?

I would start with the client request. The client can be a web app, mobile app, or Postman. The diagram shows requests such as GET /api/products/42, POST /api/products, PUT /api/products/42, and DELETE /api/products/42.

The client sends the request over HTTPS. The reverse proxy is NGINX or IIS. It owns TLS termination and also shows compression and rate limiting. The proxy then forwards the request toward the ASP.NET Core application using the HTTPS flow shown in the diagram.

Inside the .NET 8 application, the request goes through the shown middleware order: exception handling, HSTS or HTTPS handling, logging, routing, authentication, authorization, and controller or model validation.

Authentication uses JWT Bearer. It validates the token, extracts claims, and sets the user context. Authorization is a separate step. It decides whether the authenticated caller may continue. Logging uses Serilog for request logs, error logs, and audit logs.

Routing and endpoint mapping use app.MapControllers();. Attribute routing then selects the matching controller action.

The controller exposes GET /api/products, GET /api/products/{id}, POST /api/products, PUT /api/products/{id}, and DELETE /api/products/{id}. The controller handles the HTTP request and delegates the application work to ProductService.

ProductService contains the business operations shown in the diagram. These are GetAllAsync(), GetByIdAsync(id), CreateAsync(dto), UpdateAsync(id, dto), and DeleteAsync(id).

This keeps business logic separate from HTTP handling. The controller selects the operation. ProductService performs the application work. It then calls ProductRepository when stored data is needed.

ProductRepository is inside the EF Core data access layer. It provides GetAllAsync(), GetByIdAsync(id), AddAsync(entity), UpdateAsync(entity), and DeleteAsync(entity).

The repository accesses the SQL Products DB. The diagram shows Products and Categories tables. Data returns from the database to ProductRepository, then through ProductService and the controller.

The application also shows configuration from appsettings.json, the built-in dependency injection container, IMemoryCache caching, and HttpClient for outbound calls. These are supporting infrastructure components rather than extra API endpoints.

The controller returns an IActionResult. ASP.NET Core serializes the result with System.Text.Json. The response passes back through middleware in reverse order. The reverse proxy then sends the HTTPS JSON response to the client.

The diagram shows 200 OK, 201 Created, 400 Bad Request, 404 Not Found, and 500 Internal Server Error. Problem Details follows RFC 7807 and shows fields such as type, title, status, detail, and traceId.

The main benefit is clear separation between HTTP handling, business logic, and data access. The downside is extra layers and dependency wiring. For this API, that structure makes the code easier to test, debug, and change without mixing unrelated responsibilities.

The design uses several layers, but each layer has one clear job. Controllers handle HTTP requests. ProductService handles business logic. ProductRepository and EF Core handle database access. The benefit is that one type of change is less likely to affect unrelated code. JWT authentication and authorization protect access before the product operation continues. Serilog helps record requests, errors, and audit activity. Problem Details gives errors a consistent structure. Rate limiting is shown at the reverse proxy. IMemoryCache can help with repeated data access, but caching adds extra state to manage. The downside is more configuration, dependency injection, and files. We accept that extra structure because responsibilities stay clear and the API becomes easier to test, debug, and maintain.

At a high level, I would separate this into API style, version management, and content delivery. REST is simple, resource based, and works well with normal HTTP caching. GraphQL uses one /graphql endpoint and lets clients request exact fields, but complex queries and caching need more care. The main API flow is client to API Gateway, then ASP.NET Core services and data stores, with the response returning the same way. The gateway handles routing, rate limiting, authentication, authorization, and version routing. For CDN-enabled traffic, edge caching reduces latency and origin load.

This question asks me to make three practical choices. First, I need to choose how applications ask the server for data. Second, I need a safe way to change that interface without suddenly breaking existing users. Third, I need to explain how content can reach users faster from locations close to them. The goal is a design that is simple, flexible, secure, and fast. I would follow the diagram from clients, through the gateway and .NET services, to data stores, while also explaining where CDN caching fits.

• Is the API mainly public, internal, or both?
• Do clients need fixed responses or flexible field selection?
• How often do we expect breaking API changes?
• Which responses or files are safe to cache at the CDN?

I would start with the client needs. REST is resource based and normally has multiple endpoints for related data. It uses standard HTTP behavior, is easy to learn, and is naturally cacheable. The downside is over-fetching because a fixed response may return fields a client does not need.

GraphQL uses one /graphql endpoint. The client selects the exact fields it wants. It also provides a strongly typed schema. This is useful for flexible client-driven screens. The trade-off is that complex queries can become expensive, and caching or batching needs more care.

The diagram shows three versioning choices. URL versioning puts the version in the path, for example https://api.example.com/v1/orders. It is clear and cache or CDN friendly. Header versioning uses X-API-Version: 2 and keeps the URL clean. Query-string versioning uses ?version=2 and can be simple for internal APIs.

I would version breaking changes only. I would maintain backward compatibility when possible. Older versions should be deprecated with Sunset headers, documentation, and a clear migration plan.

Client applications include web, mobile, SPA, and third-party integrations. Their HTTPS request reaches the API Gateway. The gateway owns routing, rate limiting, authentication and authorization using the JWT or OAuth 2.0 mechanisms shown, request or response transformation, version routing, and optional caching.

The gateway forwards the request to the ASP.NET Core services. The response later returns from the .NET API to the gateway and then back to the client as JSON.

The .NET API Services boundary contains REST controllers, a GraphQL endpoint, and business services. Shared concerns include validation, logging, metrics, and tracing.

The service reads or writes the data-store layer. The diagram shows relational databases such as SQL Server or Aurora, NoSQL databases such as DynamoDB or Cosmos DB, Redis or ElastiCache, and object storage such as S3 or Blob Storage. Data returns to the API before the response travels back toward the client.

A CDN is a distributed network of edge servers located closer to users. For CDN-enabled traffic, DNS resolves the user to a nearby edge. On a cache hit, the edge can serve the content directly. On a miss, it forwards toward the origin path. The response returns through the CDN and may be cached when eligible.

Good cache targets shown are static JavaScript, CSS, images, fonts, public GET responses, downloadable files, and streaming content. Benefits include lower latency, higher availability, origin offload, DDoS protection, and cost optimization.

The diagram names six examples: Amazon CloudFront, Akamai, Cloudflare, Fastly, Microsoft Azure CDN, and Google Cloud CDN. REST favors simplicity and standard caching. GraphQL favors flexible client queries. Explicit API versions protect clients but require migration work. A CDN improves delivery speed, but caching rules must match the content being served.