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# 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`

```java
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;
    }
}
```