<p>Given an <code>m x n</code> binary <code>matrix</code> filled with <code>0</code>&#39;s and <code>1</code>&#39;s, <em>find the largest square containing only</em> <code>1</code>&#39;s <em>and return its area</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/max1grid.jpg" style="width: 400px; height: 319px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/max2grid.jpg" style="width: 165px; height: 165px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>matrix[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution {
public:
    int solve(vector<vector<char>>& matrix, int i, int j){
    int n = matrix.size();
     // base case 20 oct 2023
    if( i >= n || j >= matrix[0].size()){
        return 0;
    }

    // initialise the pointer right , diagonal , down
    int right = solve( matrix, i , j+1);
    int diagonal = solve(matrix, i+1, j);
    int down = solve(matrix, i+1, j+1);

    int ans = INT_MIN;
    for(int k=0; k<n; k++){
    max(right, max(diagonal, down));
    }
    return ans;
}
    int solveMem(vector<vector<char>>& matrix, int i, int j, vector<vector<int>> &dp){
        int n = matrix.size();
        int m = matrix[0].size();

        if( i >= n || j >= m){
        return 0;
    }
    if(dp[i][j] != -1){
        return dp[i][j];
    }
    int right = solveMem( matrix, i , j+1, dp);
    int diagonal = solveMem(matrix, i+1, j, dp);
    int down = solveMem(matrix, i+1, j+1, dp);

    // int ans = INT_MAX;
    // for(int k=0; k<n; k++){
    // max(ans, dp[i][j]);
    // }

     if(matrix[i][j] == '1'){
        dp[i][j] = 1 + min(right, min(down , diagonal));
     }else{
        dp[i][j] = 0;
     }
    
    return dp[i][j];
    }

    int solveTab(vector<vector<char>> matrix){

    //    int n = matrix.size();

    int row = matrix.size();
    int col = matrix[0].size();
    vector<vector<int>> dp(row+1, vector<int> (col+1, 0));

    int maxi =0;
    for(int i = row -1; i>=0; i--){
    for(int j = col-1; j>=0; j--){
    
    int right = dp[i][j+1];
    int diagonal = dp[i+1][j];
    int down = dp[i+1][j+1];


     if(matrix[i][j] == '1'){
        dp[i][j] = 1 + min(right, min(down , diagonal));
        maxi = max(maxi, dp[i][j]);
      }
    else{  
         dp[i][j] = 0;
        }
      } 
    } 
    return maxi;
    }
    int maximalSquare(vector<vector<char>>& matrix) {
  //  int n = matrix.size();
  
       int maxi =0;
       return solveTab(matrix) * solveTab(matrix);
      // return maxi; 
    }
};
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