<p>Given an <code>n x n</code> integer matrix <code>grid</code>, return <em>the minimum sum of a <strong>falling path with non-zero shifts</strong></em>.</p>

<p>A <strong>falling path with non-zero shifts</strong> is a choice of exactly one element from each row of <code>grid</code> such that no two elements chosen in adjacent rows are in the same column.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/10/falling-grid.jpg" style="width: 244px; height: 245px;" />
<pre>
<strong>Input:</strong> grid = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> 13
<strong>Explanation:</strong> 
The possible falling paths are:
[1,5,9], [1,5,7], [1,6,7], [1,6,8],
[2,4,8], [2,4,9], [2,6,7], [2,6,8],
[3,4,8], [3,4,9], [3,5,7], [3,5,9]
The falling path with the smallest sum is&nbsp;[1,5,7], so the answer is&nbsp;13.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[7]]
<strong>Output:</strong> 7
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 200</code></li>
	<li><code>-99 &lt;= grid[i][j] &lt;= 99</code></li>
</ul>

class Solution {
  public:
  int minFallingPathSum(vector<vector<int>>& grid) {
    auto min =
      vector<vector<int>>(grid.size(), vector<int>(grid[0].size(), -1));
    for (int i = 0; i < grid[0].size(); i++) min[0][i] = grid[0][i];
    for (int i = 1; i < grid.size(); i++) {
      for (int j = 0; j < grid[i].size(); j++) {
        int min_val = INT_MAX;
        for (int k = 0; k < grid[i - 1].size(); k++) {
          if (j == k) continue;
          min_val = min_val < min[i - 1][k] ? min_val : min[i - 1][k];
        }
        min[i][j] = min_val + grid[i][j];
      }
    }
    for (auto v : min) {
      for (auto i : v) cout << i << " ";
      cout << endl;
    }
    return *min_element(min.back().begin(), min.back().end());
  }
};
19600