【leetcode】1252. Cells with Odd Values in a Matrix

Topics are as follows:

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:

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:

Input: n = 2, m = 2, indices = [[1,1],[0,0]]
Output: 0
Explanation: Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.

Constraints:

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

Problem-solving ideas: first traversal indices, calculated per row were made operational a few +1. Then traverse matrix, in accordance with i rows and j columns Matrix [i] [j] corresponding to the determined several +1 are made, and determining the parity of both can.

code show as below:

class Solution(object):
    def oddCells(self, n, m, indices):
        """
        :type n: int
        :type m: int
        :type indices: List[List[int]]
        :rtype: int
        """
        dic_row = {}
        dic_column = {}
        for (r,c) in indices:
            dic_row[r] = dic_row.setdefault(r,0) + 1
            dic_column[c] = dic_column.setdefault(c, 0) + 1
        res = 0
        for i in range(n):
            for j in range(m):
                count = dic_row.get(i,0) + dic_column.get(j,0)
                if count % 2 == 1:res += 1
        return res