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
}