<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 {
    int dp[201][201];
public:

    int helper(int i , int j , int n, int m, vector<vector<int>> &grid){
         if (i == n-1 && j == m-1){
             return grid[i][j];
         }
         if (i >= n || j>= m)return 100000;
         if (dp[i][j] != -1) return dp[i][j];
         int ans1 = grid[i][j] + helper(i+1,j, n,m,grid);
         int ans2 = grid[i][j] + helper(i,j+1,n,m,grid);
         return dp[i][j] = min(ans1, ans2);
    }

    int minPathSum(vector<vector<int>>& grid) {
        memset(dp, -1,sizeof(dp));
        return helper (0,0, grid.size(), grid[0].size(), grid);
    }
};
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