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
from collections import defaultdict
class Solution:
    def canArrange(self, arr: List[int], k: int) -> bool:
        d = defaultdict(int)
        for a in arr:
            d[a % k] += 1
        res = True
        for i in range(1, (k+1)//2):
            print(i, k)
            if d[i] != d[k-i]:
                res = False
        if d[0] % 2 != 0:
            res = False
        if k % 2 == 0 and d[k//2] % 2 != 0:
            res = False
        return res
from collections import defaultdict
class Solution:
    def canArrange(self, arr: List[int], k: int) -> bool:
        d = defaultdict(int)
        for a in arr:
            d[a % k] += 1
        res = True
        for i in range(1, (k+1)//2):
            print(i, k)
            if d[i] != d[k-i]:
                res = False
        if d[0] % 2 != 0:
            res = False
        if k % 2 == 0 and d[k//2] % 2 != 0:
            res = False
        return res