<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>

// O((MN)^3)
// MN for 1st point; MN for 2nd point; MN for traversing b/w 1st and 2nd point

// O((MN)^2)
// MN for 1st point. MN for side while traversing the second point

// O(MN)
// dp
class Solution {
public:
    int n, m;
    int dp[305][305];
    vector<vector<char>> grid;

    bool isValid(int i, int j){return i>=0 && j>=0 && i<n && j<m;}
    int calc(int x, int y){
        if(!isValid(x, y)){return 0;}
        if(grid[x][y] == '0'){
            return 0;
        }
        if(x == 0 && y == 0){return 1;}
        if(dp[x][y] != -1){return dp[x][y];}
        return dp[x][y] = 1 + min(calc(x-1, y-1), \
            min(calc(x-1, y), calc(x, y-1)));
    }
    int maximalSquare(vector<vector<char>>& grid) {
        this->grid = grid;
        n = grid.size();
        m = grid[0].size();

        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                dp[i][j] = -1;
            }
        }   
        int ans = 0;
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                ans = max(ans, calc(i, j));
            }
        }
        return ans * ans;

    }


};
23600