<p>Given a <code>m x n</code> <code>grid</code> filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.</p>

<p><strong>Note:</strong> You can only move either down or right at any point in time.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/05/minpath.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[1,3,1],[1,5,1],[4,2,1]]
<strong>Output:</strong> 7
<strong>Explanation:</strong> Because the path 1 &rarr; 3 &rarr; 1 &rarr; 1 &rarr; 1 minimizes the sum.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,2,3],[4,5,6]]
<strong>Output:</strong> 12
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 200</code></li>
</ul>

class Solution {
public:
    int dyn(vector<vector<int>>& grid, vector<vector<int>>& dp, int n, int m)
    {
        if(n==0 && m==0)
        return grid[0][0];

        if(n<0 || m<0)
        return 10000000;
        
        if(dp[n][m]!=-1)
        return dp[n][m];
        
        return dp[n][m]=min(grid[n][m]+dyn(grid,dp,n-1,m),grid[n][m]+dyn(grid,dp,n,m-1));
    }
    int minPathSum(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>> dp(n,vector<int>(m,-1));
        return dyn(grid,dp,n-1,m-1);
    }
};
14800