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 <= 1000 <= nums[i] <= 100
Solutions#
Solution 1#
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| 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]
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| 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};
}
}
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| 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};
}
};
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| 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}
}
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| 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];
}
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| 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]
}
}
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| /**
* @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];
};
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| 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};
}
}
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| /**
* 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;
}
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