1252. Cells with Odd Values in a Matrix

Description

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

My Solution

Source Code

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/**
 * @param {number} n
 * @param {number} m
 * @param {number[][]} indices
 * @return {number}
 */
let oddCells = function(n, m, indices) {
  let arr = [];
  let numOdd = 0;

  //set up array
  for (let x = 0; x < n; x++) {
    arr.push(new Array(m));
    arr[x].fill(0);
  }

  for (let l = 0; l < indices.length; l++) {
    //increment everything in the requested row
    for (let i = 0; i < m; i++) {
      let index = arr[indices[l][0]][i]++;
      console.log(++index);
      if ((index + 1) % 2 === 0) {
        numOdd--;
      } else {
        numOdd++;
      }
    }
    //increment everything in the requested column
    for (let j = 0; j < n; j++) {
      let index = arr[j][indices[l][1]]++;
      if ((index + 1) % 2 === 0) {
        numOdd--;
      } else {
        numOdd++;
      }
    }
  }

  console.log(arr);

  return numOdd;
};

Analysis

There is nothing particularly special about my solution. First we set up the array of 0s. Then we iterate over the indices array and increase each specified row/column. As we do this, we check if the current element is odd and increase the counter if it is.

The running time for setting up the array is probably O(n*m) unless JavaScript's built in function for Array.fill() runs in constant time. The running time for actually solving the problem will be O(l*m + l*n) where l is length of the indices array.