Maximum Length of Repeated Subarray
Given two integer arrays `nums1` and `nums2`, find the length of the longest subarray that appears in both arrays. Use Dynamic Programming.
Why Interviewers Ask This
Microsoft evaluates this question to assess a candidate's ability to recognize overlapping subproblems, a core trait for optimizing recursive solutions. It tests whether you can transition from naive brute-force approaches to efficient Dynamic Programming, demonstrating logical structuring and memory management skills essential for building scalable backend systems.
How to Answer This Question
1. Clarify requirements by confirming if arrays are sorted or contain negative numbers, as Microsoft values precise problem scoping. 2. Propose a brute-force solution first to establish a baseline O(N*M*K) complexity, acknowledging its inefficiency for large inputs. 3. Transition to Dynamic Programming by defining the state: dp[i][j] represents the length of the common subarray ending at nums1[i-1] and nums2[j-1]. 4. Derive the recurrence relation: if elements match, increment the previous diagonal value; otherwise, reset to zero. 5. Implement the solution with nested loops, tracking the global maximum during iteration. 6. Analyze time complexity as O(N*M) and space complexity as O(N*M), then suggest space optimization to O(min(N,M)) using a rolling array technique to show advanced optimization skills.
Key Points to Cover
- Explicitly distinguishing between subarrays (contiguous) and subsequences (non-contiguous) to avoid fundamental logic errors
- Correctly identifying the base case where mismatched elements reset the current counter to zero
- Demonstrating awareness of space complexity optimization by suggesting a rolling array technique
- Articulating the recurrence relation clearly before writing any code
- Connecting the solution's efficiency to handling large-scale data typical in enterprise environments
Sample Answer
To solve the Maximum Length of Repeated Subarray problem efficiently, I would approach this using Dynamic Programming. First, I'd clarify that we are looking for contiguous elements, not subsequences. A brute-force approach checking every possible subarray would be too slow, likely resulting in Time Limit Exceeded errors on large datasets typical in Microsoft production environments.
Instead, I propose a DP table where dp[i][j] stores the length of the longest common suffix ending at index i-1 in nums1 and j-1 in nums2. The logic is straightforward: if nums1[i-1] equals nums2[j-1], then dp[i][j] = dp[i-1][j-1] + 1. If they don't match, the continuity breaks, so dp[i][j] becomes 0. We track the maximum value encountered in this table throughout the process.
For example, given [1,2,3,2,1] and [3,2,1,4,7], when comparing the '2' at index 3 in the first array with the '2' at index 1 in the second, we look back at the previous match. Since the preceding elements also matched, we extend that length. This ensures we only count contiguous segments.
This approach reduces time complexity to O(N*M) and space to O(N*M). For further optimization, I would mention using two rows instead of a full matrix to reduce space to O(min(N,M)), which demonstrates attention to resource efficiency. This method balances clarity with performance, directly addressing the constraints of real-world applications.
Common Mistakes to Avoid
- Confusing subarray with subsequence, leading to an incorrect algorithm that allows gaps between elements
- Failing to reset the DP value to zero when elements do not match, causing inflated length calculations
- Ignoring edge cases such as empty arrays or arrays with no common elements entirely
- Not discussing space optimization, missing an opportunity to showcase deeper algorithmic insight
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