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# [1497. Check If Array Pairs Are Divisible by k](https://leetcode.com/problems/check-if-array-pairs-are-divisible-by-k)

[中文文档](/solution/1400-1499/1497.Check%20If%20Array%20Pairs%20Are%20Divisible%20by%20k/README.md)

## Description

<p>Given an array of integers <code>arr</code> of even length <code>n</code> and an integer <code>k</code>.</p>

<p>We want to divide the array into exactly <code>n / 2</code> pairs such that the sum of each pair is divisible by <code>k</code>.</p>

<p>Return <code>true</code><em> If you can find a way to do that or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,2,3,4,5,10,6,7,8,9], k = 5
<strong>Output:</strong> true
<strong>Explanation:</strong> Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,2,3,4,5,6], k = 7
<strong>Output:</strong> true
<strong>Explanation:</strong> Pairs are (1,6),(2,5) and(3,4).
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,2,3,4,5,6], k = 10
<strong>Output:</strong> false
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>arr.length == n</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>n</code> is even.</li>
	<li><code>-10<sup>9</sup> &lt;= arr[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

## Solutions

### Solution 1

```python
class Solution:
    def canArrange(self, arr: List[int], k: int) -> bool:
        cnt = Counter(x % k for x in arr)
        return cnt[0] % 2 == 0 and all(cnt[i] == cnt[k - i] for i in range(1, k))
```

```java
class Solution {
    public boolean canArrange(int[] arr, int k) {
        int[] cnt = new int[k];
        for (int x : arr) {
            ++cnt[(x % k + k) % k];
        }
        for (int i = 1; i < k; ++i) {
            if (cnt[i] != cnt[k - i]) {
                return false;
            }
        }
        return cnt[0] % 2 == 0;
    }
}
```

```cpp
class Solution {
public:
    bool canArrange(vector<int>& arr, int k) {
        vector<int> cnt(k);
        for (int& x : arr) {
            ++cnt[((x % k) + k) % k];
        }
        for (int i = 1; i < k; ++i) {
            if (cnt[i] != cnt[k - i]) {
                return false;
            }
        }
        return cnt[0] % 2 == 0;
    }
};
```

```go
func canArrange(arr []int, k int) bool {
	cnt := make([]int, k)
	for _, x := range arr {
		cnt[(x%k+k)%k]++
	}
	for i := 1; i < k; i++ {
		if cnt[i] != cnt[k-i] {
			return false
		}
	}
	return cnt[0]%2 == 0
}
```

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提交于 2024-02-21 08:52 . feat: add problem tags (#2361)

1497. Check If Array Pairs Are Divisible by k

Description

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.

Constraints:

arr.length == n
1 <= n <= 10⁵
n is even.
-10⁹ <= arr[i] <= 10⁹
1 <= k <= 10⁵

Solutions

Solution 1

class Solution:
    def canArrange(self, arr: List[int], k: int) -> bool:
        cnt = Counter(x % k for x in arr)
        return cnt[0] % 2 == 0 and all(cnt[i] == cnt[k - i] for i in range(1, k))

class Solution {
    public boolean canArrange(int[] arr, int k) {
        int[] cnt = new int[k];
        for (int x : arr) {
            ++cnt[(x % k + k) % k];
        }
        for (int i = 1; i < k; ++i) {
            if (cnt[i] != cnt[k - i]) {
                return false;
            }
        }
        return cnt[0] % 2 == 0;
    }
}

class Solution {
public:
    bool canArrange(vector<int>& arr, int k) {
        vector<int> cnt(k);
        for (int& x : arr) {
            ++cnt[((x % k) + k) % k];
        }
        for (int i = 1; i < k; ++i) {
            if (cnt[i] != cnt[k - i]) {
                return false;
            }
        }
        return cnt[0] % 2 == 0;
    }
};

func canArrange(arr []int, k int) bool {
	cnt := make([]int, k)
	for _, x := range arr {
		cnt[(x%k+k)%k]++
	}
	for i := 1; i < k; i++ {
		if cnt[i] != cnt[k-i] {
			return false
		}
	}
	return cnt[0]%2 == 0
}

Java

1

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