1497. Check If Array Pairs Are Divisible by k

1497. Check If Array Pairs Are Divisible by k (Medium)

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

Related Topics:
Array, Math, Greedy

Solution 1.

Use a map to store the frequencies of remainders and check if each remainder satisfies the requirement.

1. If there are odd number 0 remainders, return false
2. For 1 <= i < k / 2, if m[i] != m[k - i], return false.

If k is an even number, we can return false if m[k / 2] % 2 != 0 but it's not needed because if m[k / 2] is odd, one of the above conditions above must be false.

// OJ: https://leetcode.com/problems/check-if-array-pairs-are-divisible-by-k/
// Author: github.com/lzl124631x
// Time: O(N + K)
// Space: O(K)
class Solution {
public:
    bool canArrange(vector<int>& A, int k) {
        unordered_map<int, int> m;
        for (int n : A) m[(n % k + k) % k]++;
        if (m[0] % 2) return false;
        for (int i = 1; i < k / 2; ++i) {
            if (m[i] != m[k - i]) return false;
        }
        return true;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/check-if-array-pairs-are-divisible-by-k/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(K)
class Solution {
public:
    bool canArrange(vector<int>& A, int k) {
        unordered_map<int, int> m;
        for (int &n : A) {
            n = (n % k + k) % k;
            m[n]++;
        }
        for (int n : A) {
            if ((n == 0 && m[n] % 2) || (n && m[n] != m[k - n])) return false;
        }
        return true;
    }
};