Leetcode Maximal Square problem solution YASH PAL, 31 July 2024 In this Leetcode Maximal Square problem solution we have 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. Topics we are covering Toggle Problem solution in Python.Problem solution in Java.Problem solution in C++. Problem solution in Python. class Solution: def maximalSquare(self, matrix: List[List[str]]) -> int: dp=[] for i in range(len(matrix)): for j in range(len(matrix[0])): matrix[i][j]=int(matrix[i][j]) if len(matrix)==1 and len(matrix[0])==1: if matrix[0][0]==0: return 0 else: return 1 for i in range(len(matrix)): x=[] for j in range(len(matrix[0])): x.append(0) dp.append(x) side=0 for i in range(len(matrix)): for j in range(len(matrix[0])): if matrix[i][j]==1: side=1 dp[i][j]=1 for i in range(1,len(matrix)): for j in range(1,len(matrix[0])): if matrix[i][j]==1: dp[i][j]=min(dp[i-1][j-1],min(dp[i-1][j], dp[i][j-1]))+1 side=max(side,dp[i][j]) return side*side Problem solution in Java. class Solution { public int maximalSquare(char[][] matrix) { if(matrix == null || matrix.length == 0 || matrix[0].length == 0) { return 0; } int m = matrix.length; int n = matrix[0].length; int[][] dp = new int[m+1][n+1]; int res = 0; for(int i=1; i<=m; i++) { for(int j=1; j<=n; j++) { if(matrix[i-1][j-1] == '1') { dp[i][j] = Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1])) + 1; res = Math.max(res, dp[i][j]); } } } return res * res; } } Problem solution in C++. class Solution { public: int maximalSquare(vector<vector<char>>& matrix) { int n = matrix.size(), m = matrix[0].size(), res = 0; vector<vector<int>> dp(n+1, vector<int>(m+1, 0)); for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (matrix[i-1][j-1] == '1') { dp[i][j] = min({dp[i-1][j], dp[i][j-1], dp[i-1][j-1]}) + 1; res = max(res, dp[i][j]); } } } return res*res; } }; coding problems solutions