<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 n,m;
int solve(int start,int end,vector<vector<int>>&grid,vector<vector<int>>&dp)
{
    if(start==n && end==m)
    return grid[start][end];
    if(start>n || end>m)
    return INT_MAX;
    if(dp[start][end]!=-1)
    return dp[start][end];
    int down=solve(start+1,end,grid,dp);
    int right=solve(start,end+1,grid,dp);
    int val1,val2;
    if(down!=INT_MAX)
    val1=grid[start][end]+down;
    else
    val1=INT_MAX;
    if(right!=INT_MAX)
    val2=grid[start][end]+right;
    else
    val2=INT_MAX;
    return dp[start][end]=min(val1,val2);    
}
    int minPathSum(vector<vector<int>>& grid) {
        n=grid.size()-1;
        m=grid[0].size()-1;
        vector<vector<int>>dp(n+1,vector<int>(m+1,-1));
        return solve(0,0,grid,dp);
    }
};
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