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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

  • 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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