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
Solution 1: Simulation
Time complexity: O((n+m)*k + n*m)
Space complexity: O(n*m)
C++
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// Author: Huahua class Solution { public: int oddCells(int n, int m, vector<vector<int>>& indices) { vector<vector<int>> a(n, vector<int>(m)); for (const auto& idx : indices) { for (int x = 0; x < m; ++x) ++a[idx[0]][x]; for (int y = 0; y < n; ++y) ++a[y][idx[1]]; } int ans = 0; for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) ans += a[i][j] & 1; return ans; } }; |
Solution 2: Counting
For each row and column, compute how many times it will be increased (odd or even).
For each a[i][j], check how many times the i-th row and j-th column were increased, if the sum is odd then a[i][j] will odd.
Time complexity: O(n*m + k)
Space complexity: O(n+m)
C++
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// Author: Huahua class Solution { public: int oddCells(int n, int m, vector<vector<int>>& indices) { vector<int> cols(m); vector<int> rows(n); for (const auto& idx : indices) { rows[idx[0]] ^= 1; cols[idx[1]] ^= 1; } int ans = 0; for (int i = 0; i < n; ++i) for (int j = 0; j < m; ++j) ans += rows[i] ^ cols[j]; return ans; } }; |
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