<p>You are given a <strong>0-indexed</strong> <code>m x n</code> binary matrix <code>grid</code>.</p>

<p>A <strong>0-indexed</strong> <code>m x n</code> difference matrix <code>diff</code> is created with the following procedure:</p>

<ul>
	<li>Let the number of ones in the <code>i<sup>th</sup></code> row be <code>onesRow<sub>i</sub></code>.</li>
	<li>Let the number of ones in the <code>j<sup>th</sup></code> column be <code>onesCol<sub>j</sub></code>.</li>
	<li>Let the number of zeros in the <code>i<sup>th</sup></code> row be <code>zerosRow<sub>i</sub></code>.</li>
	<li>Let the number of zeros in the <code>j<sup>th</sup></code> column be <code>zerosCol<sub>j</sub></code>.</li>
	<li><code>diff[i][j] = onesRow<sub>i</sub> + onesCol<sub>j</sub> - zerosRow<sub>i</sub> - zerosCol<sub>j</sub></code></li>
</ul>

<p>Return <em>the difference matrix </em><code>diff</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2022/11/06/image-20221106171729-5.png" style="width: 400px; height: 208px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1],[1,0,1],[0,0,1]]
<strong>Output:</strong> [[0,0,4],[0,0,4],[-2,-2,2]]
<strong>Explanation:</strong>
- diff[0][0] = <code>onesRow<sub>0</sub> + onesCol<sub>0</sub> - zerosRow<sub>0</sub> - zerosCol<sub>0</sub></code> = 2 + 1 - 1 - 2 = 0 
- diff[0][1] = <code>onesRow<sub>0</sub> + onesCol<sub>1</sub> - zerosRow<sub>0</sub> - zerosCol<sub>1</sub></code> = 2 + 1 - 1 - 2 = 0 
- diff[0][2] = <code>onesRow<sub>0</sub> + onesCol<sub>2</sub> - zerosRow<sub>0</sub> - zerosCol<sub>2</sub></code> = 2 + 3 - 1 - 0 = 4 
- diff[1][0] = <code>onesRow<sub>1</sub> + onesCol<sub>0</sub> - zerosRow<sub>1</sub> - zerosCol<sub>0</sub></code> = 2 + 1 - 1 - 2 = 0 
- diff[1][1] = <code>onesRow<sub>1</sub> + onesCol<sub>1</sub> - zerosRow<sub>1</sub> - zerosCol<sub>1</sub></code> = 2 + 1 - 1 - 2 = 0 
- diff[1][2] = <code>onesRow<sub>1</sub> + onesCol<sub>2</sub> - zerosRow<sub>1</sub> - zerosCol<sub>2</sub></code> = 2 + 3 - 1 - 0 = 4 
- diff[2][0] = <code>onesRow<sub>2</sub> + onesCol<sub>0</sub> - zerosRow<sub>2</sub> - zerosCol<sub>0</sub></code> = 1 + 1 - 2 - 2 = -2
- diff[2][1] = <code>onesRow<sub>2</sub> + onesCol<sub>1</sub> - zerosRow<sub>2</sub> - zerosCol<sub>1</sub></code> = 1 + 1 - 2 - 2 = -2
- diff[2][2] = <code>onesRow<sub>2</sub> + onesCol<sub>2</sub> - zerosRow<sub>2</sub> - zerosCol<sub>2</sub></code> = 1 + 3 - 2 - 0 = 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<img src="https://assets.leetcode.com/uploads/2022/11/06/image-20221106171747-6.png" style="width: 358px; height: 150px;" />
<pre>
<strong>Input:</strong> grid = [[1,1,1],[1,1,1]]
<strong>Output:</strong> [[5,5,5],[5,5,5]]
<strong>Explanation:</strong>
- diff[0][0] = onesRow<sub>0</sub> + onesCol<sub>0</sub> - zerosRow<sub>0</sub> - zerosCol<sub>0</sub> = 3 + 2 - 0 - 0 = 5
- diff[0][1] = onesRow<sub>0</sub> + onesCol<sub>1</sub> - zerosRow<sub>0</sub> - zerosCol<sub>1</sub> = 3 + 2 - 0 - 0 = 5
- diff[0][2] = onesRow<sub>0</sub> + onesCol<sub>2</sub> - zerosRow<sub>0</sub> - zerosCol<sub>2</sub> = 3 + 2 - 0 - 0 = 5
- diff[1][0] = onesRow<sub>1</sub> + onesCol<sub>0</sub> - zerosRow<sub>1</sub> - zerosCol<sub>0</sub> = 3 + 2 - 0 - 0 = 5
- diff[1][1] = onesRow<sub>1</sub> + onesCol<sub>1</sub> - zerosRow<sub>1</sub> - zerosCol<sub>1</sub> = 3 + 2 - 0 - 0 = 5
- diff[1][2] = onesRow<sub>1</sub> + onesCol<sub>2</sub> - zerosRow<sub>1</sub> - zerosCol<sub>2</sub> = 3 + 2 - 0 - 0 = 5
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution:
    def onesMinusZeros(self, grid: List[List[int]]) -> List[List[int]]:
        one_per_col = collections.Counter()

        m = len(grid)
        n = len(grid[0])
        ans = []

        for i, row in enumerate(grid):
            _no_of_one_row = sum(row)
            _no_of_zero_row = m-_no_of_one_row
            ans.append([])
            for j, col in enumerate(row):
                one_per_col[j] += col
                ans[i].append(_no_of_one_row-_no_of_zero_row)
        
        for j in one_per_col.keys():
            for i, row in enumerate(ans):
                ans[i][j] += one_per_col[j] - (n-one_per_col[j])
        
        return ans



# oneRow + zeroRow = m
# zeroRow = m-oneRow
# oneRow + oneCol - zerosRow - zerosCol
# oneRow - zerosRow + oneCol - zerosCol
# 2* oneRow - m + 2 * oneCol - n
# 2 * (oneRow + oneCol) - m -n

1292