1497. Check If Array Pairs Are Divisible by k

Description

Given an array of integers arr of even length n and an integer k.

We want to divide the array into exactly n / 2 pairs such that the sum of each pair is divisible by k.

Return true If you can find a way to do that or false otherwise.

 

Example 1:

Input: arr = [1,2,3,4,5,10,6,7,8,9], k = 5
Output: true
Explanation: Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10).

Example 2:

Input: arr = [1,2,3,4,5,6], k = 7
Output: true
Explanation: Pairs are (1,6),(2,5) and(3,4).

Example 3:

Input: arr = [1,2,3,4,5,6], k = 10
Output: false
Explanation: You can try all possible pairs to see that there is no way to divide arr into 3 pairs each with sum divisible by 10.

 

Constraints:

• arr.length == n
• 1 <= n <= 105
• n is even.
• -109 <= arr[i] <= 109
• 1 <= k <= 105

 

Solutions

Solution: Counting

• Time complexity: O(n)
• Space complexity: O(k)

 

JavaScript

/**
 * @param {number[]} arr
 * @param {number} k
 * @return {boolean}
 */
const canArrange = function (arr, k) {
  const remainderMap = new Map();

  for (const num of arr) {
    const remainder = ((num % k) + k) % k;
    const count = remainderMap.get(remainder) ?? 0;

    remainderMap.set(remainder, count + 1);
  }

  if (remainderMap.get(0) % 2) return false;

  for (let num = 1; num <= k / 2; num++) {
    const count = remainderMap.get(num);
    const pair = k - num;

    if (count !== remainderMap.get(pair)) return false;
  }
  return true;
};