LeetCode //C - 1252. Cells with Odd Values in a Matrix

1252. Cells with Odd Values in a Matrix

There is an m x n matrix that is initialized to all 0's. There is also a 2D array indices where each i n d i c e s i = r i , c i indicesi = r_i, c_i indicesi=ri,ci represents a 0-indexed location to perform some increment operations on the matrix.

For each location indicesi, do both of the following:

1. Increment all the cells on row r i r_i ri.
2. Increment all the cells on column c i c_i 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 <= 50
• 1 <= indices.length <= 100
• 0 < = r i < m 0 <= r_i < m 0<=ri<m
• 0 < = c i < n 0 <= c_i < n 0<=ci<n

From: LeetCode

Link: 1252. Cells with Odd Values in a Matrix

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Solution:

Ideas:

• Instead of updating the whole matrix, only track how many times each row and column is incremented.
• Use two arrays:
  • rowsi = number of increments on row i
  • colsj = number of increments on column j
• For each operation r, c, do:
  • rowsr++
  • colsc++
• A cell (i, j) is incremented rowsi + colsj times.
• If (rowsi + colsj) is odd, that cell is odd.
• Count all such cells.

Code:

int oddCells(int m, int n, int** indices, int indicesSize, int* indicesColSize) {
    int rows[50] = {0};
    int cols[50] = {0};

    for (int i = 0; i < indicesSize; i++) {
        rows[indices[i][0]]++;
        cols[indices[i][1]]++;
    }

    int ans = 0;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if ((rows[i] + cols[j]) % 2 == 1) {
                ans++;
            }
        }
    }

    return ans;
}