Maximal Square LeetCode Solution

Problem – Maximal Square LeetCode Solution

Given an m x n binary matrix filled with 0‘s and 1‘s, find the largest square containing only 1‘s and return its area.

Example 1:

Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 4

Example 2:

Input: matrix = [["0","1"],["1","0"]]
Output: 1

Example 3:

Input: matrix = [["0"]]
Output: 0

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 300
• matrix[i][j] is '0' or '1'.

Maximal Square LeetCode Solution in Python

class Solution:
    def maximalSquare(self, matrix: List[List[str]]) -> int:
        if matrix is None or len(matrix) < 1:
            return 0
        
        rows = len(matrix)
        cols = len(matrix[0])
        
        dp = [[0]*(cols+1) for _ in range(rows+1)]
        max_side = 0
        
        for r in range(rows):
            for c in range(cols):
                if matrix[r][c] == '1':
                    dp[r+1][c+1] = min(dp[r][c], dp[r+1][c], dp[r][c+1]) + 1 # Be careful of the indexing since dp grid has additional row and column
                    max_side = max(max_side, dp[r+1][c+1])
                
        return max_side * max_side

Maximal Square LeetCode Solution in C++

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        if (matrix.empty()) {
            return 0;
        }
        int m = matrix.size(), n = matrix[0].size(), sz = 0, pre;
        vector<int> cur(n, 0);
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int temp = cur[j];
                if (!i || !j || matrix[i][j] == '0') {
                    cur[j] = matrix[i][j] - '0';
                } else {
                    cur[j] = min(pre, min(cur[j], cur[j - 1])) + 1;
                }
                sz = max(cur[j], sz);
                pre = temp;
            }
        }
        return sz * sz;
    }
};

Maximal Square LeetCode Solution in Java

public int maximalSquare(char[][] a) {
    if(a.length == 0) return 0;
    int m = a.length, n = a[0].length, result = 0;
    int[][] b = new int[m+1][n+1];
    for (int i = 1 ; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if(a[i-1][j-1] == '1') {
                b[i][j] = Math.min(Math.min(b[i][j-1] , b[i-1][j-1]), b[i-1][j]) + 1;
                result = Math.max(b[i][j], result); // update result
            }
        }
    }
    return result*result;
}

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