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:
- Increment all the cells on row
ri. - Increment all the cells on column
ci.
Given m, n, and indices, return the number of odd-valued cells in the matrix after applying the increment to all locations in indices.
Example 1:
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:
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 <= 501 <= indices.length <= 1000 <= ri < m0 <= ci < n
Follow up: Could you solve this in O(n + m + indices.length) time with only O(n + m) extra space?
class Solution:
def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
g = [[0] * n for _ in range(m)]
for r, c in indices:
for i in range(m):
g[i][c] += 1
for j in range(n):
g[r][j] += 1
return sum(v % 2 for row in g for v in row)class Solution:
def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
row = [0] * m
col = [0] * n
for r, c in indices:
row[r] += 1
col[c] += 1
return sum((i + j) % 2 for i in row for j in col)class Solution:
def oddCells(self, m: int, n: int, indices: List[List[int]]) -> int:
row = [0] * m
col = [0] * n
for r, c in indices:
row[r] += 1
col[c] += 1
cnt1 = sum(v % 2 for v in row)
cnt2 = sum(v % 2 for v in col)
return cnt1 * (n - cnt2) + cnt2 * (m - cnt1)class Solution {
public int oddCells(int m, int n, int[][] indices) {
int[][] g = new int[m][n];
for (int[] e : indices) {
int r = e[0], c = e[1];
for (int i = 0; i < m; ++i) {
g[i][c]++;
}
for (int j = 0; j < n; ++j) {
g[r][j]++;
}
}
int ans = 0;
for (int[] row : g) {
for (int v : row) {
ans += v % 2;
}
}
return ans;
}
}class Solution {
public int oddCells(int m, int n, int[][] indices) {
int[] row = new int[m];
int[] col = new int[n];
for (int[] e : indices) {
int r = e[0], c = e[1];
row[r]++;
col[c]++;
}
int ans = 0;
for (int i : row) {
for (int j : col) {
ans += (i + j) % 2;
}
}
return ans;
}
}class Solution {
public int oddCells(int m, int n, int[][] indices) {
int[] row = new int[m];
int[] col = new int[n];
for (int[] e : indices) {
int r = e[0], c = e[1];
row[r]++;
col[c]++;
}
int cnt1 = 0, cnt2 = 0;
for (int v : row) {
cnt1 += v % 2;
}
for (int v : col) {
cnt2 += v % 2;
}
return cnt1 * (n - cnt2) + cnt2 * (m - cnt1);
}
}class Solution {
public:
int oddCells(int m, int n, vector<vector<int>>& indices) {
vector<vector<int>> g(m, vector<int>(n));
for (auto& e : indices) {
int r = e[0], c = e[1];
for (int i = 0; i < m; ++i) ++g[i][c];
for (int j = 0; j < n; ++j) ++g[r][j];
}
int ans = 0;
for (auto& row : g)
for (int v : row) ans += v % 2;
return ans;
}
};class Solution {
public:
int oddCells(int m, int n, vector<vector<int>>& indices) {
vector<int> row(m);
vector<int> col(n);
for (auto& e : indices)
{
int r = e[0], c = e[1];
row[r]++;
col[c]++;
}
int ans = 0;
for (int i : row) for (int j : col) ans += (i + j) % 2;
return ans;
}
};class Solution {
public:
int oddCells(int m, int n, vector<vector<int>>& indices) {
vector<int> row(m);
vector<int> col(n);
for (auto& e : indices)
{
int r = e[0], c = e[1];
row[r]++;
col[c]++;
}
int cnt1 = 0, cnt2 = 0;
for (int v : row) cnt1 += v % 2;
for (int v : col) cnt2 += v % 2;
return cnt1 * (n - cnt2) + cnt2 * (m - cnt1);
}
};func oddCells(m int, n int, indices [][]int) int {
g := make([][]int, m)
for i := range g {
g[i] = make([]int, n)
}
for _, e := range indices {
r, c := e[0], e[1]
for i := 0; i < m; i++ {
g[i][c]++
}
for j := 0; j < n; j++ {
g[r][j]++
}
}
ans := 0
for _, row := range g {
for _, v := range row {
ans += v % 2
}
}
return ans
}func oddCells(m int, n int, indices [][]int) int {
row := make([]int, m)
col := make([]int, n)
for _, e := range indices {
r, c := e[0], e[1]
row[r]++
col[c]++
}
ans := 0
for _, i := range row {
for _, j := range col {
ans += (i + j) % 2
}
}
return ans
}func oddCells(m int, n int, indices [][]int) int {
row := make([]int, m)
col := make([]int, n)
for _, e := range indices {
r, c := e[0], e[1]
row[r]++
col[c]++
}
cnt1, cnt2 := 0, 0
for _, v := range row {
cnt1 += v % 2
}
for _, v := range col {
cnt2 += v % 2
}
return cnt1*(n-cnt2) + cnt2*(m-cnt1)
}