<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();
        int j, numberSwaps = 0;
        int currentPosition, desiredValue;
        vector<int> zeros(n, 0);

        //count trailing zeros
        for(int i = 0; i < n; i++)
        {
            j = n - 1;
            while(j >= 0 && !grid[i][j])
            {
                zeros[i]++;
                j--;
            }
        }

        //sort using trailing zeros as reference
        for(int i = 0; i < n; i++)
        {
            currentPosition = i;
            desiredValue = n - 1 - i;
            while(currentPosition < n && zeros[currentPosition] < desiredValue)
                currentPosition++;
            if(currentPosition == n)
                return -1;
            numberSwaps += (currentPosition - i);
            while(currentPosition > i)
            {
                zeros[currentPosition] = zeros[currentPosition - 1];
                currentPosition -= 1;
            }
        }

        return numberSwaps;
    }
};
28600