[Solved] There is an m x n matrix that is initialized to all 0’s. There is also a 2D array indices where each indices[i] = [ri, ci] represents a 0-indexed location to perform some increment operations on the matrix.

Question

There is an m x n matrix that is initialized to all 0‘s. There is also a 2D array indices where each indices[i] = [ri, ci] represents a 0-indexed location to perform some increment operations on the matrix.

For each location indices[i], do both of the following:

  1. Increment all the cells on row ri.
  2. Increment all the cells on column ci.

Given mn, and indices, return the number of odd-valued cells in the matrix after applying the increment to all locations in indices.

Example 1:

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Input: m = 2, n = 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 is [[1,3,1],[1,3,1]], which contains 6 odd numbers.

Example 2:

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

Constraints:

  • 1 <= m, n <= 50
  • 1 <= indices.length <= 100
  • 0 <= ri < m
  • 0 <= ci < n

Follow up: Could you solve this in O(n + m + indices.length) time with only O(n + m) extra space?

Python Solution

class Solution:
    def oddCells(self, n: int, m: int, indices: List[List[int]]) -> int:
        rows = [0]*n
        cols = [0]*m

        for i,j in indices:
            rows[i] = rows[i]^1
            cols[j] = cols[j]^1

        c=0
        for i in range(len(rows)):
            for j in range(len(cols)):
                if rows[i]^cols[j]==1:
                    c+=1
        return c
Abhishek Sharma
Abhishek Sharma

Started my Data Science journey in my 2nd year of college and since then continuously into it because of the magical powers of ML and continuously doing projects in almost every domain of AI like ML, DL, CV, NLP.

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