题目描述

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.

样例

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

算法

开一个长度为k的数组cnt，来统计每个数对k取模后结果相同的个数。对k取模后为0的需要单独考虑，只能取模后为0的数字相互组合，所以个数必须是偶数。对于取模后大于0的需要考虑k的奇偶性，如果k是奇数，那么k-1是偶数，也就是取模后值为1的和k-1组合，2和k - 2组合，依此类推，所以只需要检查每队对应的个数是否相同；如果k是偶数，那么最后会剩下k / 2的自己配对，所以需要检查k/2的个数必须是偶数。

时间复杂度$O(n)$，空间复杂度$O(k)$。

C++ 代码

class Solution {
public:
    bool canArrange(vector<int>& arr, int k) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        vector<int> cnt(k, 0);
        for (const auto & e : arr) ++cnt[(e % k + k) % k];

        if (cnt[0] & 1) return false;
        int half = k >> 1;
        if (k & 1) {
            for (int i = 1; i <= half; ++i) {
                if (cnt[i] != cnt[k - i]) return false;
            }
        }
        else {
            for (int i = 1; i < half; ++i) {
                if (cnt[i] != cnt[k - i]) return false;
            }
            if (cnt[half] & 1) return false;
        }

        return true;
    }
};