README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1400-1499/1497.Check%20If%20Array%20Pairs%20Are%20Divisible%20by%20k/README_EN.md	1787	Weekly Contest 195 Q2	Array Hash Table Counting

1497. Check If Array Pairs Are Divisible by k

中文文档

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

Python3

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))

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

JavaScript

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;
}