Given an n x n binary grid, in one step you can choose two adjacent rows of the grid and swap them.

A grid is said to be valid if all the cells above the main diagonal are zeros.

Return the minimum number of steps needed to make the grid valid, or -1 if the grid cannot be valid.

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

Example 1:

Input: grid = [[0,0,1],[1,1,0],[1,0,0]]
Output: 3


Example 2:

Input: grid = [[0,1,1,0],[0,1,1,0],[0,1,1,0],[0,1,1,0]]
Output: -1
Explanation: All rows are similar, swaps have no effect on the grid.


Example 3:

Input: grid = [[1,0,0],[1,1,0],[1,1,1]]
Output: 0


Constraints:

• n == grid.length
• n == grid[i].length
• 1 <= n <= 200
• grid[i][j] is 0 or 1

## Solution: Bubble Sort

Counting how many tailing zeros each row has.
Then input
[0, 0, 1]
[1, 1, 0]
[1, 0, 0]
becomes [0, 1, 2]
For i-th row, it needs n – i – 1 tailing zeros.
We need to find the first row that has at least n – i – 1 tailing zeros and bubbling it up to the i-th row. This process is very similar to bubble sort.
[0, 1, 2] -> [0, 2, 1] -> [2, 0, 1]
[2, 0, 1] -> [2, 1, 0]
Total 3 swaps.

Time complexity: O(n)
Space complexity: O(n)

## C++

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