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

Solution 1

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

Java Code
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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;
    }
}

C++ Code
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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;
    }
};

Go Code
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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
}