LeetCode 1295. Find Numbers with Even Number of Digits

Description

https://leetcode.com/problems/cells-with-odd-values-in-a-matrix/

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

Explanation

Simply implement as the problem description says.

Python Solution

class Solution:
    def oddCells(self, n: int, m: int, indices: List[List[int]]) -> int:
        count = 0
        
        matrix = [[0 for j in range(m)] for i in range(n)]
        
        for indice in indices:
            r = indice[0]
            c = indice[1]
            for i in range(len(matrix)):
                for j in range(len(matrix[0])):
                    if i == r:
                        matrix[i][j] += 1
                    if j == c:
                        matrix[i][j] += 1

                        
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                if matrix[i][j] % 2 != 0:
                    count += 1

        return count
  • Time Complexity: O(N^2)
  • Space Complexity: O(N^2)

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