Question

2206. Divide Array Into Equal Pairs

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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

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

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;
  }
};

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
}