<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 minPathSum(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();

        vector<int> prev(n, -1);
        vector<int> curr(n, -1);
        curr[n-1] = grid[m-1][n-1];

        for(int i=m-1; i>=0; i--){
            for(int j=n-1; j>=0; j--){
                if(i == m-1 && j==n-1) continue;
                int bottom = (i+1 < m)? ( prev[j]) : INT_MAX;
                int right = (j+1 < n)? (curr[j+1]) : INT_MAX;

                curr[j] = grid[i][j] + min(bottom, right);
            }
            prev = curr;
        }

        return curr[0];
    }
};
12500