Maximal Square
Given an $m \times n$ binary matrix, find the largest square containing only 1's and return its area. Use Dynamic Programming.
Why Interviewers Ask This
Adobe asks this problem to evaluate a candidate's ability to recognize overlapping subproblems and apply dynamic programming for optimization. They specifically test if you can transform a brute-force geometric check into an efficient O(mn) solution, demonstrating strong algorithmic thinking and spatial reasoning skills essential for their graphics and software engineering roles.
How to Answer This Question
1. Clarify the input constraints and edge cases immediately, such as empty matrices or all-zero grids. 2. Propose a brute-force approach first to show baseline understanding, then explicitly state its inefficiency (O(m^3)). 3. Introduce the DP state definition: dp[i][j] represents the side length of the largest square ending at position (i, j). 4. Derive the recurrence relation by explaining that a cell forms a larger square only if its top, left, and top-left neighbors are also part of squares. 5. Walk through a small matrix example on a virtual whiteboard to visualize how values propagate. 6. Conclude with time and space complexity analysis, suggesting space optimization to O(n) if asked.
Key Points to Cover
- Demonstrating the transition from brute force to dynamic programming logic
- Correctly identifying the recurrence relation involving min(top, left, top_left)
- Handling boundary conditions and empty inputs gracefully
- Clearly articulating O(mn) time and O(mn) space complexity
- Using a concrete visual walkthrough to explain the state transitions
Sample Answer
To solve the Maximal Square problem efficiently, I would start by checking for edge cases like an empty matrix, returning zero immediately if found. A naive approach would check every possible square starting from each cell, resulting in cubic time complexity, which is too slow for large inputs typical at Adobe. Instead, I propose using Dynamic Programming. We define a 2D array where dp[i][j] stores the side length of the largest square whose bottom-right corner is at matrix[i][j]. The key insight is that if matrix[i][j] is '1', the size of the square ending here depends on the smallest square ending at its three neighbors: top, left, and top-left. Specifically, dp[i][j] equals one plus the minimum of those three neighbors. If any neighbor is zero, we cannot extend a larger square. By iterating through the matrix once, filling the DP table, and tracking the maximum value seen, we achieve O(mn) time complexity. For example, in a grid where a 2x2 block exists, the bottom-right cell will calculate a value of 2, reflecting the square's side length. Finally, we return the square of this maximum side length as the area. This approach balances readability with optimal performance, aligning well with Adobe's focus on clean, scalable code.
Common Mistakes to Avoid
- Failing to initialize the first row and column correctly in the DP table
- Confusing the returned value as the side length instead of the calculated area
- Attempting to use recursion without memoization, leading to Time Limit Exceeded errors
- Neglecting to check if the current cell is '0' before applying the recurrence relation
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