<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) {
    const int n = grid.size();
    int ans = 0;
    // suffixZeros[i] := the number of suffix zeros in the i-th row
    vector<int> suffixZeros;

    for (const vector<int> row : grid) {
      const auto itLastOne = find(row.rbegin(), row.rend(), 1);
      const int suffixZeroCount = distance(row.rbegin(), itLastOne);
      suffixZeros.push_back(suffixZeroCount);
    }

    for (int i = 0; i < n; ++i) {
      const int neededZeros = n - 1 - i;
      // Get the first row with suffix zeros >= `neededZeros` in
      // suffixZeros[i:..n).
      const auto it = find_if(suffixZeros.begin() + i, suffixZeros.end(),
                              [&](int count) { return count >= neededZeros; });
      if (it == suffixZeros.end())
        return -1;
      const int j = distance(suffixZeros.begin(), it);
      // Move the rows[j] to the rows[i].
      for (int k = j; k > i; --k)
        suffixZeros[k] = suffixZeros[k - 1];
      ans += j - i;
    }

    return ans;
  }
};
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