# 961. N-Repeated Element in Size 2N Array

## Description

You are given an integer array nums with the following properties:

• nums.length == 2 * n.
• nums contains n + 1 unique elements.
• Exactly one element of nums is repeated n times.

Return the element that is repeated n times.

Example 1:

Input: nums = [1,2,3,3]
Output: 3


Example 2:

Input: nums = [2,1,2,5,3,2]
Output: 2


Example 3:

Input: nums = [5,1,5,2,5,3,5,4]
Output: 5


Constraints:

• 2 <= n <= 5000
• nums.length == 2 * n
• 0 <= nums[i] <= 104
• nums contains n + 1 unique elements and one of them is repeated exactly n times.

## Solutions

• class Solution {
public int repeatedNTimes(int[] nums) {
Set<Integer> s = new HashSet<>(nums.length / 2 + 1);
for (int i = 0;; ++i) {
return nums[i];
}
}
}
}

• class Solution {
public:
int repeatedNTimes(vector<int>& nums) {
unordered_set<int> s;
for (int i = 0;; ++i) {
if (s.count(nums[i])) {
return nums[i];
}
s.insert(nums[i]);
}
}
};

• class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
s = set()
for x in nums:
if x in s:
return x


• func repeatedNTimes(nums []int) int {
s := map[int]bool{}
for i := 0; ; i++ {
if s[nums[i]] {
return nums[i]
}
s[nums[i]] = true
}
}

• function repeatedNTimes(nums: number[]): number {
const s: Set<number> = new Set();
for (const x of nums) {
if (s.has(x)) {
return x;
}
}
}


• /**
* @param {number[]} nums
* @return {number}
*/
var repeatedNTimes = function (nums) {
const s = new Set();
for (const x of nums) {
if (s.has(x)) {
return x;
}