← Bank of America Interview Insights
Classic DP problem but I still fumbled the recurrence for a second.
Use dynamic programming to compute the size of the largest square ending at each cell, then return the square of the maximum size. Explain the recurrence dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1 when matrix[i][j] == 1, and handle edge cases.
Pro tip: In banking, emphasize the O(m*n) time and O(n) space optimization, and discuss how this scales for large transaction matrices. Also mention that you validate inputs and consider memory constraints.
Confirm the definition of a square (contiguous 1s), input format (binary matrix), and expected output (area, not side length). Ask about constraints on matrix size.
Mention brute force O(m^2 n^2) and then propose dynamic programming for O(m*n) time. Explain why DP is better for large matrices.
Define dp[i][j] as the side length of the largest square ending at (i,j). If matrix[i][j] == 1, dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1; else 0.
Trace the DP on a small matrix (e.g., 3x3) to show how the recurrence works and how the maximum side length is tracked.
State time O(m*n) and space O(m*n) for 2D DP, then mention space optimization to O(n) using a 1D array or rolling rows.
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