Question

1497. Check If Array Pairs Are Divisible by k

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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 1

class Solution:
  def canArrange(self, arr: List[int], k: int) -> bool:
    cnt = Counter(x % k for x in arr)
    return cnt[0] % 2 == 0 and all(cnt[i] == cnt[k - i] for i in range(1, k))

class Solution {
  public boolean canArrange(int[] arr, int k) {
    int[] cnt = new int[k];
    for (int x : arr) {
      ++cnt[(x % k + k) % k];
    }
    for (int i = 1; i < k; ++i) {
      if (cnt[i] != cnt[k - i]) {
        return false;
      }
    }
    return cnt[0] % 2 == 0;
  }
}

class Solution {
public:
  bool canArrange(vector<int>& arr, int k) {
    vector<int> cnt(k);
    for (int& x : arr) {
      ++cnt[((x % k) + k) % k];
    }
    for (int i = 1; i < k; ++i) {
      if (cnt[i] != cnt[k - i]) {
        return false;
      }
    }
    return cnt[0] % 2 == 0;
  }
};

func canArrange(arr []int, k int) bool {
  cnt := make([]int, k)
  for _, x := range arr {
    cnt[(x%k+k)%k]++
  }
  for i := 1; i < k; i++ {
    if cnt[i] != cnt[k-i] {
      return false
    }
  }
  return cnt[0]%2 == 0
}