Microsoft·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Microsoft SWE coding round, one question the whole time. Sliding window problem that maps pretty cleanly to a known LC problem, so if you've done your prep it's manageable.

Questions Asked (1)

You're given a binary array where 1 means land and 0 means ocean, plus an integer representing how many 0s you can flip to 1. Find the maximum length of contiguous land you can create using at most that many flips.

Algorithms & Data Structures

Author's notes

Recognized this pretty fast as a sliding window setup.

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AI HintsAI Generated

Suggested Approach

Use a sliding window (two-pointer) technique to find the longest subarray containing at most K zeros. Expand the right pointer, count zeros, and when zeros exceed K, shrink the window from the left until zeros ≤ K. Track the maximum window length throughout.

Pro tip: Clarify edge cases upfront (e.g., K ≥ total zeros, all ones, empty array) and mention that the window size only increases, so you can avoid shrinking below the current max. This shows attention to detail and optimization.

1. Understand the problem and constraints

Restate the problem: find the longest contiguous subarray with at most K zeros. Ask clarifying questions about input size, K value, and expected output.

2. Choose the sliding window approach

Explain that a brute-force check of all subarrays is O(n^2), but a sliding window can solve it in O(n) time and O(1) space.

3. Initialize pointers and zero count

Set left = 0, max_len = 0, and zero_count = 0. Iterate right from 0 to n-1, incrementing zero_count when encountering a 0.

4. Expand and contract the window

When zero_count > K, move left forward until zero_count ≤ K, decrementing zero_count if the element at left is 0. Update max_len with the current window size.

5. Return the maximum length

After the loop, return max_len. Optionally, discuss how to modify the algorithm to return the actual subarray if needed.

Key Points to Mention

Sliding window technique for O(n) time complexity
Handling edge cases: K=0, K ≥ number of zeros, empty array
Space complexity O(1) with two pointers
Comparison with brute-force O(n^2) approach
Proof of correctness: window always contains at most K zeros
Potential follow-up: return the indices of the longest subarray

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