<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 fun(vector<vector<int>>& grid,int i,int j,vector<vector<int>>&dp)
{
    int m=grid.size();
    int n=grid[0].size();
    if(dp[i][j]!=-1)
    return dp[i][j];
if(i==m-1&&j==n-1)
return grid[i][j];

   int l=INT_MAX,d=INT_MAX;
    if(j<n-1)
    l=fun(grid,i,j+1,dp)+grid[i][j];
    if(i<m-1)
    d=fun(grid,i+1,j,dp)+grid[i][j];
    return dp[i][j]=min(l,d);
}
int fun(vector<vector<int>>& grid)
{
    vector<vector<int>>dp(grid.size()+1,vector<int>(grid[0].size()+1,0));
    int m=grid.size();
    int n=grid[0].size();
    dp[0][0]=grid[0][0];
    for(int j=1;j<n;j++)
    {
        dp[0][j]=dp[0][j-1]+grid[0][j];
    }
        for(int j=1;j<m;j++)
    {
        dp[j][0]=dp[j-1][0]+grid[j][0];
    }
    for(int i=1;i<m;i++)
    {
        for(int j=1;j<n;j++)
        {
          dp[i][j]=min(dp[i][j-1],dp[i-1][j])+grid[i][j];
        }
    }
    return dp[m-1][n-1];
}
    int minPathSum(vector<vector<int>>& grid) {
        vector<vector<int>>dp(grid.size()+1,vector<int>(grid[0].size()+1,-1));
        return fun(grid);
    }
};
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