# 1748. Sum of Unique Elements

## Description

You are given an integer array nums. The unique elements of an array are the elements that appear exactly once in the array.

Return the sum of all the unique elements of nums.

Example 1:

Input: nums = [1,2,3,2]
Output: 4
Explanation: The unique elements are [1,3], and the sum is 4.


Example 2:

Input: nums = [1,1,1,1,1]
Output: 0
Explanation: There are no unique elements, and the sum is 0.


Example 3:

Input: nums = [1,2,3,4,5]
Output: 15
Explanation: The unique elements are [1,2,3,4,5], and the sum is 15.


Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100

## Solutions

• class Solution {
public int sumOfUnique(int[] nums) {
int[] cnt = new int[101];
for (int x : nums) {
++cnt[x];
}
int ans = 0;
for (int x = 0; x < 101; ++x) {
if (cnt[x] == 1) {
ans += x;
}
}
return ans;
}
}

• class Solution {
public:
int sumOfUnique(vector<int>& nums) {
int cnt[101]{};
for (int& x : nums) {
++cnt[x];
}
int ans = 0;
for (int x = 0; x < 101; ++x) {
if (cnt[x] == 1) {
ans += x;
}
}
return ans;
}
};

• class Solution:
def sumOfUnique(self, nums: List[int]) -> int:
cnt = Counter(nums)
return sum(x for x, v in cnt.items() if v == 1)


• func sumOfUnique(nums []int) (ans int) {
cnt := [101]int{}
for _, x := range nums {
cnt[x]++
}
for x := 0; x < 101; x++ {
if cnt[x] == 1 {
ans += x
}
}
return
}

• function sumOfUnique(nums: number[]): number {
const cnt = new Array(101).fill(0);
for (const x of nums) {
++cnt[x];
}
let ans = 0;
for (let x = 0; x < 101; ++x) {
if (cnt[x] == 1) {
ans += x;
}
}
return ans;
}


• class Solution {
/**
* @param Integer[] $nums * @return Integer */ function sumOfUnique($nums) {
$sum = 0; for ($i = 0; $i < count($nums); $i++) {$hashtable[$nums[$i]] += 1;
if ($hashtable[$nums[$i]] == 1) {$sum += $nums[$i];
}
if ($hashtable[$nums[$i]] == 2) {$sum -= $nums[$i];
}
}
return \$sum;
}
}


• impl Solution {
pub fn sum_of_unique(nums: Vec<i32>) -> i32 {
let mut cnt = [0; 101];
for x in nums {
cnt[x as usize] += 1;
}
let mut ans = 0;
for x in 1..101 {
if cnt[x] == 1 {
ans += x;
}
}
ans as i32
}
}