# 2206. Divide Array Into Equal Pairs

## Description

You are given an integer array nums consisting of 2 * n integers.

You need to divide nums into n pairs such that:

• Each element belongs to exactly one pair.
• The elements present in a pair are equal.

Return true if nums can be divided into n pairs, otherwise return false.

Example 1:

Input: nums = [3,2,3,2,2,2]
Output: true
Explanation:
There are 6 elements in nums, so they should be divided into 6 / 2 = 3 pairs.
If nums is divided into the pairs (2, 2), (3, 3), and (2, 2), it will satisfy all the conditions.


Example 2:

Input: nums = [1,2,3,4]
Output: false
Explanation:
There is no way to divide nums into 4 / 2 = 2 pairs such that the pairs satisfy every condition.


Constraints:

• nums.length == 2 * n
• 1 <= n <= 500
• 1 <= nums[i] <= 500

## Solutions

The first step is to count the number of times each number appears in the array. Since the question asks for pairs of numbers that are part of putting two equal elements together, in other words to see if each number occurs an even number of times.

• class Solution {
public boolean divideArray(int[] nums) {
int[] cnt = new int[510];
for (int v : nums) {
++cnt[v];
}
for (int v : cnt) {
if (v % 2 != 0) {
return false;
}
}
return true;
}
}

• class Solution {
public:
bool divideArray(vector<int>& nums) {
vector<int> cnt(510);
for (int& v : nums) ++cnt[v];
for (int& v : cnt)
if (v % 2)
return false;
return true;
}
};

• class Solution:
def divideArray(self, nums: List[int]) -> bool:
cnt = Counter(nums)
return all(v % 2 == 0 for v in cnt.values())


• func divideArray(nums []int) bool {
cnt := make([]int, 510)
for _, v := range nums {
cnt[v]++
}
for _, v := range cnt {
if v%2 == 1 {
return false
}
}
return true
}