Last updated on October 10th, 2024 at 12:49 am

Here, We see Maximal Square LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

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Maximal Square LeetCode Solution

Problem Statement

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: (fig-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: (fig-2)
Input: matrix = [["0","1"],["1","0"]]
Output: 1

Example 3:
Input: matrix = [["0"]]
Output: 0

1. Maximal Square Leetcode Solution 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;
        vector<vector<int>> dp(m, vector<int>(n, 0));
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (!i || !j || matrix[i][j] == '0') {
                    dp[i][j] = matrix[i][j] - '0';
                } else {
                    dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
                }
                sz = max(dp[i][j], sz);
            }
        }
        return sz * sz;
    }
};

1.1 Explanation

1. Initialization:
   • Check if the matrix is empty. If so, return 0.
   • Initialize variables for the matrix dimensions (m and n) and the size of the largest square (sz).
   • Create a 2D DP array dp with the same dimensions as the matrix, initialized to 0.
2. DP Array Update:
   • Iterate through each cell in the matrix:
     • If the cell is on the border or contains ‘0’, set the DP value to the cell value.
     • Otherwise, calculate the size of the square ending at that cell using the minimum value from the top, left, and top-left diagonal cells in the DP array, then add 1.
   • Update sz with the maximum value in the DP array.
3. Return Result:
   • Return the area of the largest square (sz * sz).

1.2 Time Complexity

• O(m×n), where m is the number of rows and n is the number of columns.

1.3 Space Complexity

• O(m×n) for the DP array.

2. Maximal Square Leetcode Solution Java

class Solution {
    public int maximalSquare(char[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
    return 0;
      
  int max = 0, n = matrix.length, m = matrix[0].length;
  int[][] dp = new int[n + 1][m + 1];
  
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; 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;
        max = Math.max(max, dp[i][j]);
      }
    }
  }
  return max * max;
    }
}

2.1 Explanation

1. Initialization:
   • Check if the matrix is null or empty. If so, return 0.
   • Initialize variables for the matrix dimensions (n and m) and the size of the largest square (max).
   • Create a 2D DP array dp with dimensions (n+1) x (m+1), initialized to 0.
2. DP Array Update:
   • Iterate through each cell in the matrix starting from (1, 1) to (n, m):
     • If the cell contains ‘1’, calculate the size of the square ending at that cell using the minimum value from the top, left, and top-left diagonal cells in the DP array, then add 1.
     • Update max with the maximum value in the DP array.
3. Return Result:
   • Return the area of the largest square (max * max).

2.2 Time Complexity

• O(n×m), where n is the number of rows and m is the number of columns.

2.3 Space Complexity

• O(n×m) for the DP array.

3. Maximal Square Leetcode Solution JavaScript

var maximalSquare = function(matrix) {
    let max = 0
  for (let i = 0; i < matrix.length; i++) {
    for (let j = 0; j < matrix[0].length; j++) {
      if (matrix[i][j] === "0") continue
      if(i > 0 && j > 0)
        matrix[i][j] = Math.min(matrix[i - 1][j], matrix[i][j - 1], matrix[i - 1][j - 1]) + 1
      max = Math.max(matrix[i][j], max)
    }
  }
  return max ** 2
};

3.1 Explanation

1. Initialization:
   • Initialize the maximum size of the square (max) to 0.
2. Matrix Update:
   • Iterate through each cell in the matrix:
     • Skip cells with ‘0’.
     • If not on the border, calculate the size of the square ending at that cell using the minimum value from the top, left, and top-left diagonal cells, then add 1.
   • Update max with the maximum value in the matrix.
3. Return Result:
   • Return the area of the largest square (max ** 2).

3.2 Time Complexity

• O(n×m), where n is the number of rows and m is the number of columns.

3.3 Space Complexity

• O(1) since the matrix itself is used for DP.

4. Maximal Square Leetcode Solution Python

class Solution(object):
    def maximalSquare(self, matrix):
        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

4.1 Explanation

1. Initialization:
   • Check if the matrix is null or empty. If so, return 0.
   • Initialize variables for the number of rows (rows) and columns (cols).
   • Create a 2D DP array dp with dimensions (rows+1) x (cols+1), initialized to 0.
2. DP Array Update:
   • Iterate through each cell in the matrix:
     • If the cell contains ‘1’, calculate the size of the square ending at that cell using the minimum value from the top, left, and top-left diagonal cells in the DP array, then add 1.
     • Update max_side with the maximum value in the DP array.
3. Return Result:
   • Return the area of the largest square (max_side * max_side).

4.2 Time Complexity

• O(n×m), where n is the number of rows and m is the number of columns.

4.3 Space Complexity

• O(n×m) for the DP array.