README_EN.md

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

Python3

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

Java

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

C++

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

Go

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

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