m which are the dimensions of a matrix initialized by zeros and given an array
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
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.
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.
1 <= n <= 50
1 <= m <= 50
1 <= indices.length <= 100
0 <= indices[i] < n
0 <= indices[i] < m
Simply implement as the problem description says.
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 c = indice for i in range(len(matrix)): for j in range(len(matrix)): 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)): if matrix[i][j] % 2 != 0: count += 1 return count
- Time Complexity: O(N^2)
- Space Complexity: O(N^2)