<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) {
        vector<int> list;

        for(auto row : grid){
            int zeros = 0;
            for(int j = row.size()-1; j >= 0; j--){
                if(row[j] != 0)
                    break;
                zeros++;
            }
            list.push_back(zeros);
        }

        int swaps = 0;
        int j = 0;
        int n = list.size();
        int need = n-1;

        while(j < n){
            if(list[j] >= need){
                need--;
                j++;
                continue;
            }

            //Find the best close needed
            int i = j;
            while(i < n && list[i] < need)i++;

            if(i >= n)return -1;

            while(i > j){
                swap(list[i], list[i-1]);
                i--;
                swaps++;
            }

            j++;
            need--;
        }
        return swaps;
    }
};
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