<p>Given an <code>n x n</code> binary <code>grid</code>, in one step you can choose two <strong>adjacent rows</strong> of the grid and swap them.</p>

<p>A grid is said to be <strong>valid</strong> if all the cells above the main diagonal are <strong>zeros</strong>.</p>

<p>Return <em>the minimum number of steps</em> needed to make the grid valid, or <strong>-1</strong> if the grid cannot be valid.</p>

<p>The main diagonal of a grid is the diagonal that starts at cell <code>(1, 1)</code> and ends at cell <code>(n, n)</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/28/fw.jpg" style="width: 750px; height: 141px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,1],[1,1,0],[1,0,0]]
<strong>Output:</strong> 3
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/16/e2.jpg" style="width: 270px; height: 270px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,1,1,0],[0,1,1,0],[0,1,1,0]]
<strong>Output:</strong> -1
<strong>Explanation:</strong> All rows are similar, swaps have no effect on the grid.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/16/e3.jpg" style="width: 200px; height: 200px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[1,1,0],[1,1,1]]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>n == grid.length</code> <code>== grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 200</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code></li>
</ul>

class Solution {
public:
    int minSwaps(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<int> v(n, 0);
        for (int i = 0; i < n; i++) {
            int cnt = 0;
            for (int j = 0; j < n; j++) {
                if (grid[i][j])
                    cnt = 0;
                else
                    cnt++;
            }
            v[i] = cnt;
        }
        int cnt = n, ans = 0;
        for (int i = 0; i < n - 1; i++) {
            cnt--;
            int prev = 0;
            for (int j = 0; j < n; j++) {
                int temp = v[j];
                v[j] = prev;
                prev = temp;
                if (temp >= cnt) {
                    ans += j - i;
                    break;
                }
                if (j == n - 1)
                    return -1;
            }
        }
        return ans;
    }
};
28700