<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 getSmallestNeighbor(vector<vector<int>> & grid, int y, int x) {
        if (y - 1 < 0 && x - 1 < 0)
            return 0;
        else if (y - 1 < 0)
            return grid[y][x-1];
        else if (x - 1 < 0) {
            return grid[y-1][x];
        }

        return min(grid[y-1][x], grid[y][x-1]);
    }

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

        for (int y = 0; y < n; ++y) {
            for (int x = 0; x < m; ++x) {
                grid[y][x] += getSmallestNeighbor(grid, y, x);
            }
        }

        return grid[n - 1][m - 1];
    }
};
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