Welcome to Subscribe On Youtube

2341. Maximum Number of Pairs in Array

Description

You are given a 0-indexed integer array nums. In one operation, you may do the following:

  • Choose two integers in nums that are equal.
  • Remove both integers from nums, forming a pair.

The operation is done on nums as many times as possible.

Return a 0-indexed integer array answer of size 2 where answer[0] is the number of pairs that are formed and answer[1] is the number of leftover integers in nums after doing the operation as many times as possible.

 

Example 1:

Input: nums = [1,3,2,1,3,2,2]
Output: [3,1]
Explanation:
Form a pair with nums[0] and nums[3] and remove them from nums. Now, nums = [3,2,3,2,2].
Form a pair with nums[0] and nums[2] and remove them from nums. Now, nums = [2,2,2].
Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [2].
No more pairs can be formed. A total of 3 pairs have been formed, and there is 1 number leftover in nums.

Example 2:

Input: nums = [1,1]
Output: [1,0]
Explanation: Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [].
No more pairs can be formed. A total of 1 pair has been formed, and there are 0 numbers leftover in nums.

Example 3:

Input: nums = [0]
Output: [0,1]
Explanation: No pairs can be formed, and there is 1 number leftover in nums.

 

Constraints:

  • 1 <= nums.length <= 100
  • 0 <= nums[i] <= 100

Solutions

  • class Solution {
        public int[] numberOfPairs(int[] nums) {
            int[] cnt = new int[101];
            for (int x : nums) {
                ++cnt[x];
            }
            int s = 0;
            for (int v : cnt) {
                s += v / 2;
            }
            return new int[] {s, nums.length - s * 2};
        }
    }
    
  • class Solution {
    public:
        vector<int> numberOfPairs(vector<int>& nums) {
            vector<int> cnt(101);
            for (int& x : nums) {
                ++cnt[x];
            }
            int s = 0;
            for (int& v : cnt) {
                s += v >> 1;
            }
            return {s, (int) nums.size() - s * 2};
        }
    };
    
  • class Solution:
        def numberOfPairs(self, nums: List[int]) -> List[int]:
            cnt = Counter(nums)
            s = sum(v // 2 for v in cnt.values())
            return [s, len(nums) - s * 2]
    
    
  • func numberOfPairs(nums []int) []int {
    	cnt := [101]int{}
    	for _, x := range nums {
    		cnt[x]++
    	}
    	s := 0
    	for _, v := range cnt {
    		s += v / 2
    	}
    	return []int{s, len(nums) - s*2}
    }
    
  • function numberOfPairs(nums: number[]): number[] {
        const n = nums.length;
        const count = new Array(101).fill(0);
        for (const num of nums) {
            count[num]++;
        }
        const sum = count.reduce((r, v) => r + (v >> 1), 0);
        return [sum, n - sum * 2];
    }
    
    
  • /**
     * @param {number[]} nums
     * @return {number[]}
     */
    var numberOfPairs = function (nums) {
        const cnt = new Array(101).fill(0);
        for (const x of nums) {
            ++cnt[x];
        }
        const s = cnt.reduce((a, b) => a + (b >> 1), 0);
        return [s, nums.length - s * 2];
    };
    
    
  • public class Solution {
        public int[] NumberOfPairs(int[] nums) {
            int[] cnt = new int[101];
            foreach(int x in nums) {
                ++cnt[x];
            }
            int s = 0;
            foreach(int v in cnt) {
                s += v / 2;
            }
            return new int[] {s, nums.Length - s * 2};
        }
    }
    
    
  • impl Solution {
        pub fn number_of_pairs(nums: Vec<i32>) -> Vec<i32> {
            let n = nums.len();
            let mut count = [0; 101];
            for &v in nums.iter() {
                count[v as usize] += 1;
            }
            let mut sum = 0;
            for v in count.iter() {
                sum += v >> 1;
            }
            vec![sum as i32, (n - sum * 2) as i32]
        }
    }
    
    

All Problems

All Solutions