Meta Python Developer Interview Questions & Answers

I would scan the values once while keeping the current maximum, one selected index, and the number of maximum values seen. A larger value resets the selected index and count. A tied value replaces the selected index with probability one divided by the updated count. This gives every maximum index an equal chance and uses constant extra space.

The practical solution is to apply reservoir sampling only to indexes whose values equal the largest value seen so far. During one scan, keep the current maximum, the selected index, the number of tied maximums seen, and a flag that records whether the iterable contained an item.

1. Should I focus on Python language behavior, or also explain the runtime and standard library?
2. Which Python version and execution environment should I assume?
3. Would you like a small code example together with production tradeoffs and edge cases?

The first value becomes the current maximum. When a later value is larger, it becomes the new maximum, its index becomes the selected index, and the count resets to one. When a value equals the current maximum, increase the count and replace the selected index with probability one divided by that count. In Python, calling randrange with the count and checking whether the result is zero gives exactly that probability.

For values [4, 9, 2, 9, 9], indexes 1, 3, and 4 each have probability one third. The function works with lists, generators, and other one pass iterables because it never reads an item twice. Empty input raises ValueError. Values must support consistent greater than and equality comparisons. Floating point NaN needs an explicit policy because its normal comparisons do not define a usable maximum. The algorithm takes linear time and constant extra memory.

The function uses enumerate to obtain each value and its zero based index during one iteration. The first item initializes the current maximum, selected index, and maximum count. A larger value replaces the current maximum and resets the count to one because all earlier candidates are no longer maximum values. An equal value increases the count. The new index replaces the previous selection only when randrange returns zero. After k tied maximum values have been processed, each matching index has probability one divided by k of being selected. Only a fixed number of variables are stored, so the extra memory cost remains constant.

A common mistake is storing every maximum index in a list. That is correct for selection fairness but can use linear extra memory. Another mistake is replacing the chosen index with probability one half for every tie. That makes later indexes more likely. Candidates may also forget to reset the count when a larger value appears, scan the iterable twice, return the maximum value instead of its index, or ignore empty input. Another mistake is assuming NaN follows ordinary maximum comparison rules.

State the invariant clearly. After processing any prefix with k occurrences of its maximum value, each of those k indexes has probability one divided by k of being selected. Then explain the reset for a larger value, the replacement rule for a tie, and the linear time with constant extra memory.

I would use breadth first search with collections.deque because every allowed move has the same cost. I would keep a predecessor dictionary that records which cell discovered each new cell and also acts as the visited set. When the target is reached, I would follow those links backward, reverse the result, and return both the number of cells in the shortest path and the path coordinates.

Use breadth first search because every allowed move has the same cost. Assume zero means open, one means blocked, movement is allowed in eight directions, and distance counts the cells in the returned path. Python deque supports efficient removal from the left, so cells are processed in first in first out order. The first time breadth first search discovers a cell, it has found a shortest route to that cell. Store every discovered coordinate in a predecessor dictionary. Its value is the coordinate that led to it. Dictionary membership also marks the cell as visited, so each cell enters the queue only once. When the target is reached, follow predecessor links backward, reverse the collected coordinates, and return the path length with the path. Return negative one and an empty list for an empty matrix, blocked endpoints, or an unreachable target. The function raises ValueError for rows with different lengths. For r rows and c columns, time is O(r times c), since each cell checks at most eight neighbors. Memory is O(r times c) for the queue, predecessor dictionary, and returned path. The function does not modify or copy the matrix. Weighted moves require a weighted shortest path algorithm.

1. Should I focus on Python language behavior, or also explain the runtime and standard library?
2. Which Python version and execution environment should I assume?
3. Would you like a small code example together with production tradeoffs and edge cases?

The function treats zero as open and one as blocked. It allows horizontal, vertical, and diagonal movement. It first handles an empty matrix and verifies that every row has the same length. It returns negative one with an empty path when the start or target is blocked. A deque processes coordinates in breadth first order. The predecessor dictionary maps each discovered cell to the cell that discovered it. Membership in that dictionary prevents duplicate visits. Once the target is removed from the queue, the function follows predecessor links back to the start, reverses the collected coordinates, and returns len(path) as the distance. The matrix is only read and is not copied or modified. For the included matrix, the result is distance three with the path [(0, 0), (1, 1), (2, 2)].

Common mistakes include using a list with pop(0), which shifts all remaining items, marking a cell visited only after removing it from the queue, forgetting diagonal movement, changing the input matrix without stating that behavior, counting moves while the code returns cells, and failing to check blocked endpoints. Another mistake is storing only distances, because distances alone do not provide the predecessor links needed to reconstruct the route. A candidate may also overwrite a predecessor after a cell was already discovered, which can add duplicate work and make the path logic harder to reason about.

State the assumptions first. Explain that equal move costs make breadth first search correct. Then show how deque provides efficient queue behavior, how the predecessor dictionary also acts as the visited set, and how following parent links reconstructs the path. Finish with O(r times c) time and O(r times c) memory.

I would build a prefix sum list once. Each position stores the number of ones before that position. Then getOne returns prefix[end_idx + 1] minus prefix[start_idx], so every valid query takes O(1) time. Building the prefix list takes O(n) time and O(n) extra memory.

The practical solution is to preprocess the binary list into a prefix sum list. The prefix list has one extra leading zero. Each later value stores the total number of ones seen so far. For the input [1, 0, 1, 1, 0], the prefix list is [0, 1, 1, 2, 3, 3]. To count ones from index 1 through index 3, compute prefix[4] minus prefix[1]. The result is 3 minus 1, which is 2.

1. Should I focus on Python language behavior, or also explain the runtime and standard library?
2. Which Python version and execution environment should I assume?
3. Would you like a small code example together with production tradeoffs and edge cases?

Python list indexing takes constant time, so each query performs two lookups and one subtraction. Preprocessing visits every input value once, so it takes O(n) time. The prefix list contains n plus one integers, so it uses O(n) extra memory.

The implementation should reject noninteger indexes, negative indexes, indexes outside the list, and ranges where start_idx is greater than end_idx. It should also validate that each input item is exactly the integer 0 or 1. The counter stores a snapshot of the totals, so later changes to the original list do not update the counter. This design is best for fixed data with many queries. Frequent value changes would require rebuilding the prefix data.

The RangeOneCounter constructor validates that every input item is exactly the integer 0 or 1. It then creates a prefix sum list with one extra leading zero. For each input item, it appends the previous total plus the current value. The getOne method validates both indexes and returns prefix[end_idx + 1] minus prefix[start_idx]. Construction takes O(n) time and O(n) extra memory. Each valid getOne call takes O(1) time. The stored totals are independent of later changes to the original input list.

A common mistake is summing a slice during every query. That makes each query take time proportional to the requested range and also allocates a new list for the slice. Another mistake is forgetting that end_idx is inclusive, which causes the wrong prefix position to be used. Candidates may also forget the leading zero, accidentally allow Python negative indexes, accept Boolean or floating point values as binary integers, fail to reject start_idx greater than end_idx, or assume later changes to the original list automatically update the stored prefix totals.

State the tradeoff first. Spend O(n) time and O(n) memory once, then answer every valid inclusive range query in O(1) time. Show the formula prefix[end_idx + 1] minus prefix[start_idx], explain the leading zero, and mention that updates require rebuilding the prefix data.

I would use binary search and return the index of any local minimum. I compare the middle value with each neighbor that exists. If it is smaller than both existing neighbors, I return its index. If the left neighbor is smaller, I search the left half. Otherwise, I search the right half. A boundary value is compared with only its existing neighbor. The solution takes O(log n) time and O(1) extra space.

The practical solution is to return the index of any local minimum with binary search. I assume the list is not empty and adjacent values are different. A value is a local minimum when it is smaller than every neighbor that exists. Therefore, the first and last positions need only one comparison. A list with one value has a local minimum at index zero.

1. Should I focus on Python language behavior, or also explain the runtime and standard library?
2. Which Python version and execution environment should I assume?
3. Would you like a small code example together with production tradeoffs and edge cases?

The function keeps inclusive left and right indexes. On each loop, it calculates the middle index without creating a new list. It checks the left neighbor only when the middle index is greater than zero. It checks the right neighbor only when the middle index is less than the final index. These conditions prevent invalid access.

If the middle value is smaller than both existing neighbors, the function returns it. If the left neighbor is smaller, a local minimum exists somewhere from the current left boundary through the left neighbor, so the search moves left. Otherwise, the search moves right. Each loop removes about half of the remaining indexes.

For [9, 7, 3, 5, 8], the function returns index 2, whose value is 3. The time cost is O(log n), and the extra memory cost is O(1).

The function returns the index of any strict local minimum. It raises ValueError for an empty sequence because no valid index exists. It keeps inclusive left and right boundaries and examines the middle position. A missing neighbor is treated as satisfying that side of the local minimum test, which safely handles index zero, the final index, and a sequence with one value. If the middle value is smaller than every existing neighbor, the function returns its index. If the left neighbor is smaller than the middle value, the function continues in the left half. Otherwise, it continues in the right half. For [9, 7, 3, 5, 8], it returns index 2, and the value at that index is 3.

Common mistakes include reading the left or right neighbor before checking whether that index exists. Another mistake is requiring two neighbors for the first or last value. Candidates also move the wrong search boundary after finding a smaller neighbor, which can remove the side that contains the guaranteed local minimum. List slicing is unnecessary because it creates new lists and increases memory use. Another error is ignoring the assumption that adjacent values are different. Without that assumption, equal values can make the chosen direction ambiguous.

State the strict local minimum definition and the input assumptions first. Then explain the boundary checks, the direction rule, and why each loop removes about half of the remaining indexes. Finish with O(log n) time and O(1) extra space.

I would use two pointers, one at each end of the string. While the characters match, I move both pointers inward. At the first mismatch, I try skipping either the left character or the right character. A helper checks whether the remaining range is a palindrome. If either check succeeds, I return True. This works because only one deletion is allowed. The solution takes O(n) time and O(1) auxiliary space.

The problem asks whether a string is already a palindrome or can become one after deleting at most one character. A palindrome reads the same in both directions. A two-pointer method fits because we can compare mirrored characters from the two ends. When the first mismatch appears, only the two mismatching characters can be candidates for the one deletion.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is one string named s.

The output is a Boolean value. Return True when s is already a palindrome or can become one after deleting at most one character. Otherwise, return False.

For the diagram's example, s = "abca". The expected result is True because deleting 'b' at index 1 produces "aca".

Set left to the first index and right to the last index.

Compare s[left] with s[right]. If they match, move both pointers inward.

The central invariant is that every character outside the current inclusive range [left, right] has already matched its mirrored character. At most one deletion is still available until the first mismatch is handled.

For s = "abca", the indices are 0, 1, 2, and 3.

The characters are 'a', 'b', 'c', and 'a'.

Start with left = 0 and right = 3.

The helper is_palindrome(i, j) checks whether the inclusive range s[i:j + 1] is a palindrome. It also uses two pointers and does not create a new substring.

First, compare s[0] = 'a' with s[3] = 'a'.

They match, so move inward. The new state is left = 1 and right = 2.

Next, compare s[1] = 'b' with s[2] = 'c'.

They do not match. Because only one deletion is allowed, there are two possible choices. Skip the left mismatching character or skip the right mismatching character.

The code first calls is_palindrome(2, 2). This represents skipping 'b' at index 1.

The checked range contains only 'c'. A one-character string is a palindrome, so the helper returns True.

Python stops evaluating the or expression after this successful check. The second helper call is not executed for this example. The method returns True immediately.

All mirrored pairs before the first mismatch already match and can stay in the final palindrome.

At the first mismatch, one of the two mismatching characters must be removed. Deleting a different character would leave the current mismatch unresolved.

Therefore, checking the range after skipping the left character and the range after skipping the right character covers every possible valid one-deletion repair.

If either range is a palindrome, the original string can become a palindrome after at most one deletion.

The nested helper checks one inclusive range of the original string. It returns False at the first unequal mirrored pair. If its pointers meet or cross, the range is a palindrome and it returns True.

The main loop performs the same check on the complete string. Matching characters move both pointers inward.

At the first mismatch, the method returns the result of the two possible helper checks. This also stops any later processing.

If the main loop finishes without finding a mismatch, the original string is already a palindrome, so the method returns True.

The time complexity is O(n), where n is the string length. The main scan is linear. After the first mismatch, the helper examines at most the remaining range. The total work is still proportional to n.

The auxiliary space complexity is O(1). The algorithm stores only a few pointer variables and does not copy substrings.

Relevant edge cases include an empty string, a one-character string, an existing palindrome such as "racecar", a string fixed by one deletion such as "abca", and a string such as "abc" that cannot be fixed with only one deletion.

Use two pointers to compare mirrored characters from the outside inward. The invariant is that every character outside the current inclusive range [left, right] has already matched correctly. When the current characters match, move both pointers inward. At the first mismatch, one of those two mismatching characters must be the deleted character. Check the remaining range after skipping the left character and after skipping the right character. If either range is a palindrome, return True. Otherwise, return False.

The time complexity is O(n), where n is the length of the string. The main loop compares characters from both ends. If it finds a mismatch, one or two helper checks may examine the remaining range. Even in the worst case, the total number of comparisons is only a constant multiple of n, so the time remains O(n). The auxiliary space complexity is O(1) because the algorithm uses only pointer variables and does not create copied substrings.

A common mistake is checking only one deletion choice at the first mismatch. Candidates may always skip the left character or always skip the right character, but either choice can be wrong. Another mistake is moving the pointers before saving the mismatch positions. Some solutions create sliced strings, which adds extra memory. It is also incorrect to continue processing after the helper has found a valid result or to say that both helper calls always execute, because Python's or uses short-circuit evaluation.

When you reach the first mismatch, explain why the deleted character must be one of those two mismatching characters. Then test the two inclusive ranges (left + 1, right) and (left, right - 1).

I use two pointers, one for each string, and a list to build the result. While either string still has characters, I append the next character from word1 when available, then the next character from word2 when available. This keeps the required alternating order. When one string ends, the loop continues with the remaining characters of the other string. The time complexity is O(m + n), and the auxiliary space complexity is O(m + n).

The problem gives us two strings and asks us to create one new string. We take one character from word1, then one from word2, and repeat. If one string ends first, we append the remaining characters from the longer string. Two pointers work well because each pointer tracks the next unused character in one input string.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The inputs are two strings named word1 and word2. The output is one new merged string.

In the diagram, word1 is "abcde" and word2 is "pqr". The expected output is "apbqcrde".

The characters must keep their original order inside each input string. We only alternate the string from which we take the next character.

I use two integer pointers named i and j. Pointer i tracks the next unused character in word1. Pointer j tracks the next unused character in word2.

I also use a list named result. Appending characters to a Python list is efficient. After processing both strings, I join the list into one final string.

The main invariant is that result contains every character before index i in word1 and every character before index j in word2 in the required alternating order.

Set i = 0 and j = 0. Both pointers start at the first character of their strings.

Create result = []. It is empty because no characters have been processed yet.

The loop continues while i is inside word1 or j is inside word2. The or condition is important because processing must continue after one string ends.

Start with word1 = "abcde", word2 = "pqr", i = 0, j = 0, and result = [].

Step 1: i is valid. Append word1[0], which is "a". The result becomes ["a"]. Increase i to 1.

Step 2: j is valid. Append word2[0], which is "p". The result becomes ["a", "p"]. Increase j to 1.

Step 3: append word1[1], which is "b". The result becomes ["a", "p", "b"]. Increase i to 2.

Step 4: append word2[1], which is "q". The result becomes ["a", "p", "b", "q"]. Increase j to 2.

Step 5: append word1[2], which is "c". The result becomes ["a", "p", "b", "q", "c"]. Increase i to 3.

Step 6: append word2[2], which is "r". The result becomes ["a", "p", "b", "q", "c", "r"]. Increase j to 3.

Now word2 has ended, but word1 still has two characters.

Step 7: append word1[3], which is "d". The result becomes ["a", "p", "b", "q", "c", "r", "d"]. Increase i to 4.

Step 8: append word1[4], which is "e". The result becomes ["a", "p", "b", "q", "c", "r", "d", "e"]. Increase i to 5.

Both strings are now finished. Joining the list returns "apbqcrde".

During each loop iteration, the algorithm appends the next unused character from word1 when one exists. It then appends the next unused character from word2 when one exists.

This preserves the order of characters inside both strings. It also creates the required alternating order while both strings still have characters.

When one string ends, its condition becomes false. The condition for the other string can still be true, so its remaining characters are appended in order.

Therefore, every input character is added exactly once, and the returned string is correct.

The function initializes i, j, and result. The while condition uses or so the loop continues until both strings are fully processed.

The first if statement checks whether word1 still has an unused character. If it does, the code appends word1[i] and increments i.

The second if statement checks whether word2 still has an unused character. If it does, the code appends word2[j] and increments j.

Finally, "".join(result) combines the collected characters and returns the merged string.

Let m be the length of word1 and n be the length of word2. Each input character is processed once, so the time complexity is O(m + n).

The result list can hold m + n characters, so the auxiliary space complexity is O(m + n).

The same code handles an empty string, two empty strings, strings with very different lengths, and strings that each contain one character.

Use one pointer for each string and one list for the merged output. During each loop iteration, append word1[i] if i is valid, then append word2[j] if j is valid. Increment only the pointer whose character was appended. The central invariant is that result contains all characters before i in word1 and all characters before j in word2 in the correct alternating order. The loop uses or, so it continues until both strings have been completely processed.

Let m be the length of word1 and n be the length of word2. Each character from both strings is appended exactly once, so the time complexity is O(m + n). The result list grows to contain all m + n characters before they are joined, so the auxiliary space complexity is O(m + n).

A common mistake is using and instead of or in the while condition. That stops the loop when the shorter string ends and loses the remainder of the longer string. Another mistake is appending word2[i] inside the word1 branch or incrementing the wrong pointer. Candidates may also forget the boundary checks and cause an IndexError. Another mistake is changing the original order of characters. It is also incorrect to claim O(1) auxiliary space because the result list grows with the total input size.

Before writing the loop, explain that i and j always point to the next unused characters and that result already contains all earlier characters in the correct order.

I would use iterative quickselect. First, I convert the kth largest position into the ascending index len(nums) - k. Then I partition the current inclusive range using its last value as the pivot. The pivot moves to its final sorted index. I compare that index with the target and keep only the side that can still contain the answer. The average time is O(n), the worst case is O(n²), and the auxiliary space is O(1).

The problem asks us to return the kth largest value in an integer array. We do not need to fully sort the array. Quickselect searches for one final sorted position. For an array of length n, the kth largest value belongs at ascending index n - k. Each partition places one pivot at its final sorted index and lets us discard the side that cannot contain the answer.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is an integer array nums and an integer k. The output is the kth largest value, not an index.

For nums = [3, 2, 1, 5, 6, 4] and k = 2, the length is 6. The target index in ascending order is:

target_index = len(nums) - k = 6 - 2 = 4

If the array were fully sorted as [1, 2, 3, 4, 5, 6], index 4 would contain 5. Therefore, 5 is the second largest value.

I set left = 0 and right = 5. These are inclusive boundaries, so the current search interval is [0, 5].

The main invariant is that the target index always stays inside the current interval [left, right].

The partition function uses nums[right] as the pivot. It moves values less than or equal to the pivot to the left side. Values greater than the pivot stay on the right side. It then puts the pivot into its final sorted position and returns that position as pivot_index.

The first search interval is [0, 5]. The pivot is nums[5], which is 4.

Before partitioning, nums is [3, 2, 1, 5, 6, 4]. After partitioning, nums becomes [3, 2, 1, 4, 6, 5]. The pivot 4 is now at index 3.

The target index is 4. Since pivot_index 3 is smaller than target_index 4, the answer must be on the right side. I update left to pivot_index + 1, which is 4. The new interval is [4, 5].

The current interval contains [6, 5]. The pivot is nums[5], which is 5.

After partitioning this interval, nums becomes [3, 2, 1, 4, 5, 6]. The pivot 5 is now at index 4.

Now pivot_index equals target_index. The algorithm stops and returns nums[4], which is 5. Only two partition rounds were executed.

Every partition places one pivot at the same index it would have in a fully sorted array.

If pivot_index is smaller than target_index, the target cannot be at the pivot or to its left. It must be on the right. If pivot_index is larger than target_index, the target must be on the left.

Therefore, each update removes only positions that cannot contain the answer. When pivot_index reaches target_index, nums[pivot_index] is the kth largest value.

The main function computes target_index and starts with the complete array interval. It repeatedly calls partition.

The partition function keeps a variable named store_index. This marks where the next value less than or equal to the pivot should be placed. The loop reads every value in the current interval except the pivot. When a value is less than or equal to the pivot, the code swaps it into store_index and moves store_index one position right.

After the loop, the pivot is swapped into store_index. The function returns store_index as pivot_index.

The main loop returns immediately when pivot_index equals target_index. Otherwise, it changes either left or right and repeats.

The average time is O(n). Quickselect normally removes a large part of the remaining search interval after each partition.

The worst-case time is O(n²). This can happen when the last value repeatedly creates very uneven partitions and removes only one position at a time.

The algorithm modifies nums in place and uses O(1) auxiliary space.

When k = 1, it returns the maximum value. When k = len(nums), it returns the minimum value. Duplicate values, negative values, and zero are handled correctly.

The key idea is to find one ranked position instead of sorting the complete array. The kth largest value belongs at ascending index len(nums) - k. Quickselect partitions the current inclusive interval around a pivot. After partitioning, the pivot is at its final sorted index. The central invariant is that the target index always remains inside [left, right]. If the pivot index is too small, only the right side can contain the target. If it is too large, only the left side can contain the target. The search stops when pivot_index equals target_index.

Let n be the number of values in nums. The average time is O(n) because quickselect usually reduces the remaining search interval after each partition. The worst-case time is O(n²). This happens when the chosen pivot repeatedly creates a very uneven split, so only one position is removed in each round. The implementation changes the array in place and uses only a few variables. Therefore, its auxiliary space is O(1).

A common mistake is using k directly as an ascending index instead of computing len(nums) - k. Another mistake is moving the wrong boundary after comparing pivot_index with target_index. Candidates may forget that both left and right are inclusive. In the partition function, using the wrong comparison can break duplicate handling. It is also incorrect to claim guaranteed O(n) time because the worst case is O(n²). Finally, this implementation changes the input array, which should be stated.

State the invariant before writing code: the target index always remains inside [left, right], and every completed partition places its pivot at its final sorted index.

I first sort the intervals by their start value. Then I keep a merged list and compare each next interval with the last merged interval. If the next start is less than or equal to the last end, the intervals overlap or touch, so I extend the last end. Otherwise, I append a new interval. Sorting makes possible overlaps adjacent. The total time is O(n log n), and the result uses O(n) space in the worst case.

The problem gives a list of intervals and asks us to combine all overlapping intervals. The result must cover exactly the same ranges without overlaps. The main idea is to sort the intervals by their start value. After sorting, any interval that can overlap the current merged range appears next to it, so we can build the answer with one forward traversal.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is a list of intervals. Each interval contains a start value and an end value.

The example input is [[8,10], [1,3], [15,18], [2,6]].

The expected output is [[1,6], [8,10], [15,18]]. The output covers exactly the same ranges as the input, but it is sorted and contains no overlapping intervals.

The diagram treats endpoints as inclusive. Therefore, touching intervals also merge. For example, [1,4] and [4,5] merge into [1,5].

Sort the intervals by their start value. The sorted order is [[1,3], [2,6], [8,10], [15,18]].

Create a list named merged. Put a copy of the first sorted interval into it. The initial state is [[1,3]].

The central invariant is that merged is always sorted, non-overlapping, and equal to the exact union of all intervals processed so far.

For each next interval [start, end], compare start with the end of the last interval in merged.

If start <= merged[-1][1], the intervals overlap or touch. Update the last end to max(merged[-1][1], end).

If start > merged[-1][1], there is a gap. Append [start, end] as a new merged interval.

Start with merged = [[1,3]].

Process [2,6]. The state before is [[1,3]]. Check 2 <= 3. The condition is true, so the intervals overlap. Set the last end to max(3,6), which is 6. The state becomes [[1,6]].

Process [8,10]. The state before is [[1,6]]. Check 8 <= 6. The condition is false, so there is a gap. Append [8,10]. The state becomes [[1,6], [8,10]].

Process [15,18]. The state before is [[1,6], [8,10]]. Check 15 <= 10. The condition is false, so append [15,18]. The final state is [[1,6], [8,10], [15,18]].

Sorting places possible overlaps beside each other. Because merged is already sorted and non-overlapping, a new interval only needs to be compared with the last merged interval.

When the next start is less than or equal to the last end, extending the last end preserves the full covered range. When the next start is greater than the last end, no earlier merged interval can overlap the new interval, so appending it is safe.

After every step, the invariant remains true. Therefore, the final list is sorted, non-overlapping, and covers exactly the same ranges as the input.

The function first handles an empty input by returning an empty list. It sorts the input list in place using each interval's start value.

It initializes merged with a copy of the first sorted interval. The loop then processes every remaining interval in sorted order. It reads the end of the last merged interval, checks the overlap condition, and either updates that end or appends a new interval.

After all intervals are processed, the function returns merged.

Sorting takes O(n log n) time. The merge traversal takes O(n) time. Therefore, the total time is O(n log n).

The returned merged list may contain up to n intervals, so it uses O(n) space when the output is counted, matching the diagram.

Important edge cases are an empty input, one interval, completely disjoint intervals, nested intervals such as [1,10] and [2,5], and touching intervals such as [1,4] and [4,5].

Sort all intervals by their start value. This places intervals that may overlap next to each other. Maintain a list named merged. Its invariant is that it is sorted, contains no overlaps, and represents the exact union of every interval processed so far. Compare each new interval only with the last merged interval. If the new start is less than or equal to the last end, extend the last end to the larger end value. Otherwise, append a new interval.

Let n be the number of intervals. Sorting takes O(n log n) time. The merge loop processes the intervals in one forward pass, which takes O(n) time. The total time is therefore O(n log n). The merged result may contain n intervals when none overlap, so the diagram reports O(n) space for the output. The input list itself is sorted in place.

A common mistake is forgetting to sort by the start value. Another is comparing the new interval with something other than the last merged interval. Using start < last_end instead of start <= last_end would fail to merge touching intervals under the diagram's inclusive-endpoint rule. Replacing the last end with end instead of max(last_end, end) fails for nested intervals. Candidates may also forget the empty-input case or incorrectly claim O(n) total time while ignoring sorting.

Before writing code, state the invariant clearly: merged is always sorted, non-overlapping, and covers exactly all intervals processed so far.

I would use a min heap. I place the first value from each non-empty sorted array into the heap as a tuple containing the value, array index, and element index. Each next() call removes the smallest tuple, returns its value, and pushes the next value from the same array when one exists. The heap therefore always exposes the next global minimum. Initialization is O(k), each next() call is O(log k), all N values take O(N log k), and extra space is O(k).

The class receives k arrays that are already sorted. It must return one smallest remaining value on every next() call. A min heap fits this problem because it quickly exposes the smallest current candidate. We keep only one candidate from each unfinished array, so the values are produced lazily instead of building the complete merged list first.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is a list of k sorted integer arrays.

For the example:

A0 = [1, 4, 4] A1 = [1, 3, 5] A2 = [2, 6]

The class returns one value per next() call. Consecutive calls return:

[1, 1, 2, 3, 4, 4, 5, 6]

After every value has been returned, another call to next() raises StopIteration.

Python heapq implements a min heap. This means the smallest tuple is always available at the top.

Each heap entry is:

(value, array_index, element_index)

The value controls the heap order. The array index identifies the source array. The element index identifies the value's position inside that array.

The central invariant is that the heap contains the smallest not-yet-returned value from every array that still has remaining elements.

We save the input arrays and create an empty heap.

For every non-empty array, we add its first value. For the example, the initial logical heap contents are:

(1, A0, 0), (1, A1, 0), (2, A2, 0)

In the Python code, A0, A1, and A2 are represented by array indices 0, 1, and 2. Empty arrays are skipped. We then call heapify to build the min heap.

Step 1 starts with [(1,A0,0), (1,A1,0), (2,A2,0)]. We pop (1,A0,0), so next() returns 1. The next value in A0 is A0[1] = 4, so we push (4,A0,1). The heap becomes [(1,A1,0), (2,A2,0), (4,A0,1)].

Step 2 pops (1,A1,0). We return 1. The next value in A1 is A1[1] = 3, so we push (3,A1,1). The heap becomes [(2,A2,0), (3,A1,1), (4,A0,1)].

Step 3 pops (2,A2,0). We return 2. The next value in A2 is A2[1] = 6, so we push (6,A2,1). The heap becomes [(3,A1,1), (4,A0,1), (6,A2,1)].

Step 4 pops (3,A1,1). We return 3. The next value in A1 is A1[2] = 5, so we push (5,A1,2). The heap becomes [(4,A0,1), (5,A1,2), (6,A2,1)].

Step 5 pops (4,A0,1). We return 4. The next value in A0 is A0[2] = 4, so we push (4,A0,2). The heap becomes [(4,A0,2), (5,A1,2), (6,A2,1)].

Step 6 pops (4,A0,2). We return 4. A0 has no later value, so nothing is pushed. The heap becomes [(5,A1,2), (6,A2,1)].

Step 7 pops (5,A1,2). We return 5. A1 is exhausted, so nothing is pushed. The heap becomes [(6,A2,1)].

Step 8 pops (6,A2,1). We return 6. A2 is exhausted, so nothing is pushed. The heap becomes empty.

The final sequence is [1, 1, 2, 3, 4, 4, 5, 6]. All eight input values appear exactly once in nondecreasing order.

Before every next() call, the heap contains the smallest remaining value from each unfinished array. Any later value in one of those arrays cannot be smaller than its current candidate because each array is sorted.

Therefore, the smallest value in the heap is also the smallest value remaining across all arrays. After removing it, we add the next value from the same array. This restores the invariant for the following call.

The constructor stores the arrays and builds a heap containing the first value from every non-empty array.

The next() method first checks whether the heap is empty. If it is empty, there are no remaining values, so it raises StopIteration.

Otherwise, the method removes the smallest tuple. It calculates next_index = element_index + 1. If that position exists in the same array, it pushes the successor into the heap. It then returns the popped value.

Let k be the number of input arrays and N be the total number of values.

Initialization takes O(k) time because we inspect each array, add at most one item from it, and heapify at most k entries. Each next() call performs one heappop and at most one heappush. These operations take O(log k), so returning all N values takes O(N log k) time.

The heap contains at most one entry per unfinished array, so auxiliary space is O(k).

Empty arrays are skipped. Duplicate values are handled correctly. One non-empty array is returned in its original order. Calling next() after the heap is empty raises StopIteration.

The key idea is to keep only the smallest remaining candidate from each unfinished array. A min heap is suitable because it exposes the smallest candidate in O(log k) time when it is removed. Each heap entry stores the value, its source array, and its position in that array. After removing the smallest entry, we add only the next value from the same sorted array. The invariant is that the heap always contains the smallest remaining value from every unfinished array. This produces the merged order lazily and keeps the heap size at most k.

Let k be the number of arrays and N be the total number of values. Initialization takes O(k) time because the code visits each array, stores at most one initial item from it, and heapifies at most k items. Each next() call performs one heap removal and at most one heap insertion. Each operation takes O(log k), so one next() call is O(log k). Returning all N values takes O(N log k) total time. The heap stores at most k tuples, so auxiliary space is O(k).

A common mistake is adding every value from every array to the heap. That uses O(N) extra space and removes the lazy streaming benefit. Another mistake is storing only the value, which loses the source array and position needed to find its successor. Candidates may also forget to push the next value from the same array, fail to skip empty arrays, or forget to raise StopIteration after exhaustion. It is also incorrect to claim O(1) time per next() call or O(1) auxiliary space.

State the invariant before writing code: the heap contains one smallest remaining candidate from every unfinished array. Then explain how each pop and optional push restores that invariant.

I would use linear search because the list is unsorted. I start at index 0 and compare each value with the target. If they match, I return the current index immediately. This returns the first occurrence when duplicates exist. If the loop finishes without a match, I return -1. I process each element at most once and stop when the answer is found. The worst-case time is O(n), and the auxiliary space is O(1).

The problem asks us to find the first position of a target value in an unsorted list. If the target is missing, we return -1. Because the values are not ordered, binary search cannot safely remove half of the search area. Linear search fits the problem because it checks the values from left to right and can stop as soon as it finds a match.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input contains a list of integers called nums and an integer called target. The output is the index of the first target occurrence. An index is the position of a value in the list. If the target does not occur, the function returns -1.

In the diagram, nums = [4, 1, 7, 3, 9] and target = 3. The expected output is 3 because nums[3] = 3.

The list is unsorted. This means there is no ordering that supports binary search. The direct method is linear search. We read the values from left to right and compare each one with the target.

The central invariant is: before checking index i, every earlier index from 0 through i - 1 has already been checked, and none of those earlier values equals the target.

Traversal begins at index 0, which is the leftmost position. The target is 3. At the start, no indices have been checked.

The Python code uses enumerate(nums). This provides the current index and the current value during each loop step.

At index 0, the current value is 4. We check 4 == 3. The condition is false, so we continue. The checked indices become [0]. The target has not been found yet.

At index 1, the current value is 1. We check 1 == 3. The condition is false, so we continue. The checked indices become [0, 1]. The target has not been found yet.

At index 2, the current value is 7. We check 7 == 3. The condition is false, so we continue. The checked indices become [0, 1, 2]. The target has not been found yet.

At index 3, the current value is 3. We check 3 == 3. The condition is true, so the function returns 3 and stops. The checked indices are [0, 1, 2, 3].

Index 4 is not processed because the algorithm stops immediately after finding the target. The algorithm processes 4 of the 5 elements.

All positions before index 3 were checked and did not contain the target. At index 3, the value equals the target. Therefore, index 3 is the correct returned position. Because the search moves from left to right and stops at the first match, it also returns the first occurrence when duplicates exist.

If the loop finishes without returning, every element has been checked and none equals the target. Returning -1 is then correct.

The function accepts nums and target. The for loop uses enumerate to get each index and value. The condition value == target checks whether the current value is the answer. When the condition is true, the function returns the index immediately. If the loop ends without a match, the function returns -1.

The worst-case time complexity is O(n) because the target may be absent or may appear at the last position. The best-case time is O(1) when the first element matches. The auxiliary space complexity is O(1) because the algorithm uses only a few variables and does not create another collection that grows with the input.

Relevant edge cases include an empty list, a one-element list, the target at the first position, the target being absent, duplicate target values, negative values, and zero.

The key insight is that an unsorted list gives us no safe way to skip values. We therefore scan from left to right with linear search. At each index, we compare the current value with the target. We return immediately when they match. The invariant is that before checking index i, all earlier indices have already been checked and none contains the target. This proves that the first match is the first occurrence. If no match is found after the loop, the target is absent.

Let n be the number of elements in nums. In the worst case, the algorithm checks all n elements, so the time complexity is O(n). In the best case, the first element matches, so the time is O(1). We process the input at most once and may stop early. The auxiliary space is O(1) because the algorithm stores only the current index and value. It does not build a list, set, or dictionary.

Common mistakes include returning the matching value instead of its index, continuing after a match instead of returning immediately, returning the last duplicate instead of the first occurrence, forgetting to return -1 when the target is absent, trying to use binary search on an unsorted list, and claiming that the algorithm always processes every element even though it may stop early.

Say the invariant before writing the loop: every earlier index has already been checked and does not contain the target. Then place the return directly inside the matching condition to show the early stop.