Verkada·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Verkada SWE interview with an in-place array merge problem. Pretty clean algorithmic question, nothing too wild, but the O(1) space constraint is where it gets interesting.

Questions Asked (1)

You have two sorted arrays. The first array has extra space at the end to accommodate all elements from the second. Merge the second array into the first in-place, keeping everything sorted, using constant extra space.

Algorithms & Data Structures

Author's notes

The naive approach is just copy everything into a temp array and sort it, but they want O(1) space so that's out.

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AI HintsAI Generated

Suggested Approach

Recognize that merging from the front would require shifting elements, so instead merge from the end of the first array (which has extra space) backwards. Use three pointers: one for the last valid element of the first array, one for the last element of the second array, and one for the last position of the merged array. Compare elements and place the larger one at the end, moving pointers accordingly.

Pro tip: Emphasize that this approach achieves O(m+n) time and O(1) space, and mention that it avoids the need for additional arrays or shifting. Also, clarify that the first array's extra space is at the end, so filling from the back is natural and efficient.

1. Clarify the problem and constraints

Confirm that the first array has enough space to hold all elements from both arrays, and that the extra space is at the end. Ask if the arrays are sorted in ascending order and if there are any duplicate handling requirements.

2. Choose the merging direction

Explain that merging from the front would require shifting elements, leading to O(m*n) time. Instead, merge from the end to utilize the extra space and achieve O(m+n) time with constant space.

3. Set up pointers

Initialize three pointers: i = m-1 (last valid element in first array), j = n-1 (last element in second array), and k = m+n-1 (last position in first array).

4. Iterate and merge

While j >= 0, compare nums1[i] and nums2[j]. Place the larger value at nums1[k], then decrement the corresponding pointer and k. If i < 0, copy remaining elements from nums2.

5. Handle remaining elements and verify

After the loop, if any elements remain in nums2, copy them to the beginning of nums1. If elements remain in nums1, they are already in place. Walk through a small example to verify correctness.

Key Points to Mention

Time complexity: O(m+n) where m and n are the number of valid elements in the first and second arrays respectively.
Space complexity: O(1) because we modify the first array in-place without using extra space proportional to input size.
The importance of merging from the end to avoid shifting elements and to utilize the extra space efficiently.
Edge cases: one array empty, all elements of one array smaller than the other, and handling when i becomes negative.
The algorithm is stable and preserves the relative order of equal elements if we choose to place the larger one first (or handle equals carefully).
Potential pitfalls: off-by-one errors in pointer initialization and loop conditions, and forgetting to copy remaining elements from the second array.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.