<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 solve(int i,int j,vector<vector<int>>& grid,vector<vector<int>>& dp){
    if(i==0&&j==0)return grid[0][0];
    if(i<0||j<0)return INT_MAX;
    // if(dp[i][j]!=-1)return dp[i][j];
    int up=solve(i-1,j, grid,dp);
    int left=solve(i,j-1, grid,dp);
    // return dp[i][j]=grid[i][j]+min(up,left);
    return grid[i][j]+min(up,left);

}
    int minPathSum(vector<vector<int>>& grid) {
        //recursion
        int n=grid.size();
        int m=grid[0].size();
        vector<int>dp(m,0);
        // return solve(n-1,m-1,grid,dp);

        //Tabulation
        for(int i=0;i<n;i++){
            vector<int>temp(m,0);
            for(int j=0;j<m;j++){
            
                if(i==0&&j==0)temp[j]=grid[i][j];
                else{
                int up=grid[i][j];
                if(i>0)up+=dp[j];
                else up+=1e9;
                int left=grid[i][j];
                if(j>0)left+=temp[j-1];
                else left+=1e9;
                temp[j]=min(up,left);
                }
            }
            dp=temp;
        }
        return dp[m-1];
    }
};
12800