LeetCode - Maximal Square

1 minute read

Problem description

description

Given a 2D binary matrix filled with 0’s and 1’s, find the largest square containing only 1’s and return its area.

Example:

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Input: 

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4

Analysis

This is square not rectangle. So we don’t need to use the rectangle method to do this.

The basic idea is to use DP.

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public class Solution {
    public int maximalSquare(char[][] matrix) {
        int rows = matrix.length, cols = rows > 0 ? matrix[0].length : 0;
        int[][] dp = new int[rows + 1][cols + 1];
        int maxsqlen = 0;
        for (int i = 1; i <= rows; i++) {
            for (int j = 1; j <= cols; j++) {
                if (matrix[i-1][j-1] == '1'){
                    dp[i][j] = Math.min(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
                    maxsqlen = Math.max(maxsqlen, dp[i][j]);
                }
            }
        }
        return maxsqlen * maxsqlen;
    }
}

We can also compress the 2d array into 1d array.

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public class Solution {
    public int maximalSquare(char[][] matrix) {
        int rows = matrix.length, cols = rows > 0 ? matrix[0].length : 0;
        int[] dp = new int[cols + 1];
        int maxsqlen = 0, prev = 0;
        for (int i = 1; i <= rows; i++) {
            for (int j = 1; j <= cols; j++) {
                int temp = dp[j];
                if (matrix[i - 1][j - 1] == '1') {
                    dp[j] = Math.min(Math.min(dp[j - 1], prev), dp[j]) + 1;
                    maxsqlen = Math.max(maxsqlen, dp[j]);
                } else {
                    dp[j] = 0;
                }
                prev = temp;
            }
        }
        return maxsqlen * maxsqlen;
    }
}

What to improve

  • read the problem carefully.