1497. Check If Array Pairs Are Divisible by k

https://leetcode.com/problems/check-if-array-pairs-are-divisible-by-k/

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.

Example 4:

Input: arr = [-10,10], k = 2
Output: true

Example 5:

Input: arr = [-1,1,-2,2,-3,3,-4,4], k = 3
Output: true

Constraints:

• arr.length == n

• 1 <= n <= 10^5

• n is even.

• -10^9 <= arr[i] <= 10^9

• 1 <= k <= 10^5

问给定偶数个数目的数组能否俩俩配对，每对和都是K的倍数。和是K的倍数 (a + b) % K == 0 =>取余后和mod K为0 (a % K + b % K) % K == 0 。统计余数a出现的频率，看K - a的频率是否与之相等。余数为0时满足频率为偶数。

class Solution:
    def canArrange(self, arr: List[int], k: int) -> bool:
        m = collections.defaultdict(int)
        for a in arr:
            m[a % k] += 1
        for key, value in m.items():
            if key == 0:
                if m[0] % 2 != 0: return False
                continue
            if m[k - key] != value: return False
        return True