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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

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• 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;
      }
  };
  

• 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))
  
  

• 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
  }
  

• function canArrange(arr: number[], k: number): boolean {
      const cnt = Array(k).fill(0);
  
      for (const x of arr) {
          cnt[((x % k) + k) % k]++;
      }
  
      for (let i = 1; i < k; i++) {
          if (cnt[i] !== cnt[k - i]) return false;
      }
  
      return cnt[0] % 2 === 0;
  }
  
  

• function canArrange(arr, k) {
      const cnt = Array(k).fill(0);
  
      for (const x of arr) {
          cnt[((x % k) + k) % k]++;
      }
  
      for (let i = 1; i < k; i++) {
          if (cnt[i] !== cnt[k - i]) return false;
      }
  
      return cnt[0] % 2 === 0;
  }
  
  

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