<p>Given an <code>n x n</code> array of integers <code>matrix</code>, return <em>the <strong>minimum sum</strong> of any <strong>falling path</strong> through</em> <code>matrix</code>.</p>

<p>A <strong>falling path</strong> starts at any element in the first row and chooses the element in the next row that is either directly below or diagonally left/right. Specifically, the next element from position <code>(row, col)</code> will be <code>(row + 1, col - 1)</code>, <code>(row + 1, col)</code>, or <code>(row + 1, col + 1)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/11/03/failing1-grid.jpg" style="width: 499px; height: 500px;" />
<pre>
<strong>Input:</strong> matrix = [[2,1,3],[6,5,4],[7,8,9]]
<strong>Output:</strong> 13
<strong>Explanation:</strong> There are two falling paths with a minimum sum as shown.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/11/03/failing2-grid.jpg" style="width: 164px; height: 365px;" />
<pre>
<strong>Input:</strong> matrix = [[-19,57],[-40,-5]]
<strong>Output:</strong> -59
<strong>Explanation:</strong> The falling path with a minimum sum is shown.
</pre>

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

<ul>
	<li><code>n == matrix.length == matrix[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>-100 &lt;= matrix[i][j] &lt;= 100</code></li>
</ul>

class Solution {
public:
    int getMinPath(int i,int j,vector<vector<int>>& matrix,vector<vector<bool>>&vis,
    int n,int m,vector<vector<int>>&dp){
       
        if(i<0||i>=n||j<0||j>=m||vis[i][j]){
            return INT_MAX;
        }
        if(dp[i][j]!=INT_MAX){
            return dp[i][j];
        }
        if(i==n-1){
            return dp[i][j]=matrix[i][j];
        }
        vis[i][j]=1;
        int left=getMinPath(i+1,j-1,matrix,vis,n,m,dp);
        int right=getMinPath(i+1,j+1,matrix,vis,n,m,dp);
        int dn=getMinPath(i+1,j,matrix,vis,n,m,dp);
        vis[i][j]=0;
        return dp[i][j]=min(left,min(dn,right))+matrix[i][j];
    }
    int minFallingPathSum(vector<vector<int>>& matrix) {
        int n=matrix.size();
        int m=matrix[0].size();
        vector<vector<bool>>vis(n,vector<bool>(m,0));
        vector<vector<int>>dp(n,vector<int>(m,INT_MAX));

        int ans=INT_MAX;
        for(int i=0;i<m;i++){
            ans=min(ans,getMinPath(0,i,matrix,vis,n,m,dp));
        }
        return ans;
    }
};
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