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
class Solution {
public boolean canArrange(int[] arr, int k) {
Map<Integer, Integer> map = new HashMap<>();
for (int x : arr) {
int mod = x % k;
if (mod < 0)
mod += k;
if (map.containsKey(k - mod)) {
map.put(k - mod, map.get(k - mod) - 1);
if (map.get(k - mod) == 0)
map.remove(k - mod);
} else
map.put(mod, map.getOrDefault(mod, 0) + 1);
}
if (map.containsKey(0) && map.get(0) % 2 == 0)
map.remove(0);
return map.size() == 0;
}
}