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+---
+id: maximal-square
+title: Maximal Square
+sidebar_label: 221 Maximal Square
+tags:
+- Dynamic Programming
+- Java
+- Matrix
+- Array
+description: "This document provides a solution where we find the largest square containing only 1's and return its area."
+---
+
+## Problem
+
+You are 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.
+
+### Examples
+
+**Example 1:**
+
+![image](https://github.com/vivekvardhan2810/codeharborhub.github.io/assets/91594529/4816deda-e3ba-47d4-b6d8-b38dd4fe67e2)
+
+**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:**
+
+![image](https://github.com/vivekvardhan2810/codeharborhub.github.io/assets/91594529/43f9556f-bdde-4d5c-b8ab-a5807c51cb3c)
+
+**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 $1$
+
+---
+## Approach
+There are four approaches discussed that helps to obtain the solution:
+
+1. **Dynamic Programming Table**:
+   - Use a 2D DP array **'dp'** where **'dp[i][j]'** represents the side length of the largest square whose bottom-right corner is at position **'(i, j)'**.
+     
+   - The value of **'dp[i][j]'** is determined by the values of **'dp[i-1][j]'**, **'dp[i][j-1]'**, and **'dp[i-1][j-1]'**.
+
+2. **Transition**:
+     
+   - If **'matrix[i][j]'** is $1$:
+     -  If **'i'** or **'j'** is $0$ (first row or first column), **'dp[i][j]'** is $1$ because the largest square ending there can only be of $size1$.
+     
+     -  Otherwise, **'dp[i][j]'** is the minimum of **'dp[i-1][j]'**, **'dp[i][j-1]'**, and **'dp[i-1][j-1]'** plus $1$. This is because we can form a larger square only if all three adjacent squares can also form squares of $1's$.
+        
+3. **Max Side Length**:
+   
+   - Track the maximum side length of squares found during the iteration.
+
+4. **Result**:
+     
+   - The area of the largest square is the square of the maximum side length found.
+
+## Solution for Maximal Square
+
+This problem can be solved using dynamic programming. The problem requires to Utilize a DP table where each entry represents the side length of the largest square ending at that position.
+
+#### Code in Java
+    
+ ```java
+class Solution {
+    public int maximalSquare(char[][] matrix) {
+        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
+            return 0;
+        }
+
+        int rows = matrix.length;
+        int cols = matrix[0].length;
+        int maxSide = 0;
+
+        // Create a DP array to store the size of the largest square ending at each position
+        int[][] dp = new int[rows][cols];
+
+        // Fill the DP array
+        for (int i = 0; i < rows; i++) {
+            for (int j = 0; j < cols; j++) {
+                if (matrix[i][j] == '1') {
+                    if (i == 0 || j == 0) {
+                        // If we're at the first row or first column, the largest square ending here is just 1
+                        dp[i][j] = 1;
+                    } else {
+                        // Otherwise, calculate the size of the square based on the surrounding squares
+                        dp[i][j] = Math.min(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1;
+                    }
+                    // Update the maximum side length found
+                    maxSide = Math.max(maxSide, dp[i][j]);
+                }
+            }
+        }
+
+        // The area of the largest square is side length squared
+        return maxSide * maxSide;
+    }
+}
+
+```
+
+### Complexity Analysis
+
+#### Time Complexity: O($m$ x $n$)
+
+> **Reason**: The algorithm involves iterating through each cell of the matrix once, leading to a time complexity of $𝑂(𝑚 × 𝑛)$, where $𝑚$ is the number of rows and $𝑛$ is the number of columns.
+
+#### Space Complexity: O($m$ × $n$)
+
+> **Reason**: The space complexity is $𝑂(𝑚 × 𝑛)$ due to the additional DP array used. This could be optimized to $O(n)$ by reusing a single row of DP values, but in the given solution, we use a full 2D DP array.
+
+# References
+
+- **LeetCode Problem:** [Maximal Square](https://leetcode.com/problems/maximal-square/description/)
+- **Solution Link:** [Maximal Square Solution on LeetCode](https://leetcode.com/problems/maximal-square/solutions/)
+- **Authors LeetCode Profile:** [Vivek Vardhan](https://leetcode.com/u/vivekvardhan43862/)