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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 is the number of pairs that are formed and answer 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 and nums and remove them from nums. Now, nums = [3,2,3,2,2].
Form a pair with nums and nums and remove them from nums. Now, nums = [2,2,2].
Form a pair with nums and nums and remove them from nums. Now, nums = .
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 and nums 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 = 
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;
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 := 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;
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).