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Formatted question description: https://leetcode.ca/all/2341.html

2341. Maximum Number of Pairs in Array

  • Difficulty: Easy.
  • Related Topics: Array, Hash Table, Counting.
  • Similar Questions: Sort Characters By Frequency, Top K Frequent Words, Sort Array by Increasing Frequency.

Problem

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

Solution (Java, C++, Python)

  • class Solution {
    public int[] numberOfPairs(int[] nums) {
        Arrays.sort(nums);
        int pairs = 0;
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                nums[i] = -1;
                nums[i - 1] = -1;
                pairs++;
            }
        }
        return new int[] {pairs, nums.length - (2 * pairs)};
    }
}

############

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:
    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]

############

# 2341. Maximum Number of Pairs in Array
# https://leetcode.com/problems/maximum-number-of-pairs-in-array/

class Solution:
    def numberOfPairs(self, nums: List[int]) -> List[int]:
        n = len(nums)
        removed = 0
        
        for x in Counter(nums).values():
            removed += x // 2
        
        return [removed, n - removed * 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};
    }
};

  • 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]
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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