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

Solution 1: Counting

We can count the occurrences of each number $x$ in the array $\textit{nums}$ and record them in a hash table or array $\textit{cnt}$.

Then, we traverse $\textit{cnt}$. For each number $x$, if the occurrence count $v$ of $x$ is greater than $1$, we can select two $x$'s from the array to form a pair. We divide $v$ by $2$ and take the floor value to get the number of pairs that can be formed by the current number $x$. We then add this number to the variable $s$.

The remaining count is the length of the array $\textit{nums}$ minus the number of pairs formed multiplied by $2$, i.e., $n - s \times 2$.

The answer is $[s, n - s \times 2]$.

The time complexity is $O(n)$, and the space complexity is $O(C)$. Here, $n$ is the length of the array $\textit{nums}$, and $C$ is the range of numbers in the array $\textit{nums}$, which is $101$ in this problem.

Python3JavaC++GoTypeScriptRustJavaScriptC#C

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]

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};
    }
};

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];
}

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]
    }
}

/**
 * @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};
    }
}

/**
 * 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;
}

2341. Maximum Number of Pairs in Array

Description

Solutions

Solution 1: Counting

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