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Weekly Contest 302 Q1
Array
Hash Table
Counting

2341. Maximum Number of Pairs in Array

中文文档

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

Solution 1

Python3

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]

Java

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};
    }
}

C++

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};
    }
};

Go

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}
}

TypeScript

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];
}

Rust

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]
    }
}

JavaScript

/**
 * @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];
};

C#

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};
    }
}

C

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* numberOfPairs(int* nums, int numsSize, int* returnSize) {
    int count[101] = {0};
    for (int i = 0; i < numsSize; i++) {
        count[nums[i]]++;
    }
    int sum = 0;
    for (int i = 0; i < 101; i++) {
        sum += count[i] >> 1;
    }
    int* ans = malloc(sizeof(int) * 2);
    ans[0] = sum;
    ans[1] = numsSize - sum * 2;
    *returnSize = 2;
    return ans;
}