Given n and m which are the dimensions of a matrix initialized by zeros and given an array indices where indices[i] = [ri, ci]. For each pair of [ri, ci] you have to increment all cells in row ri and column ci by 1.

Return the number of cells with odd values in the matrix after applying the increment to all indices.

Example 1:

Example 1

Input: n = 2, m = 3, indices = [[0,1],[1,1]]
Output: 6
Explanation: Initial matrix = [[0,0,0],[0,0,0]].
After applying first increment it becomes [[1,2,1],[0,1,0]].
The final matrix will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.

Example 2:

Example 2:

Constraints:

• 1 <= n <= 50
• 1 <= m <= 50
• 1 <= indices.length <= 100
• 0 <= indices[i][0] < n
• 0 <= indices[i][1] < m

Solution1:

class Solution:
    def oddCells(self, n: int, m: int, indices: List[List[int]]) -> int:
        x, y = [0] * n, [0] * m
        for i in indices:
            x[i[0]] += 1
            y[i[1]] += 1
        return sum([1 for j in y for i in x if (j+i) % 2])

We can think of row and column individually and then use list comprehension to find the summation as each cell (instead of making a matrix explicitly).
我們可以單獨(dú)考慮行和列，然后使用列表推導(dǎo)來求出目標(biāo)矩陣中元素的值（而不是顯式地創(chuàng)建矩陣）。

Solution2

class Solution:
    
    def oddCells(self, n: int, m: int, indices: List[List[int]]) -> int:
        matrix = [[0 for i in range(m)] for j in range(n)] 
        
        def inc(x, y):
            for i in range(m):
                matrix[x][i] += 1
            for i in range(n):
                matrix[i][y] += 1
                
        for ind in indices:
            inc(ind[0], ind[1])
            
        return sum([0 if n % 2 == 0 else 1 for l in matrix for n in l])

This solution is to create a matrix explicitly and update each cell of it by creating a function.
該解決方案是顯式創(chuàng)建矩陣并通過創(chuàng)建一個(gè)函數(shù)來更新矩陣中的每個(gè)值。