In a array A
of size 2N
, there are N+1
unique elements, and exactly one of these elements is repeated N
times.
Return the element repeated N
times.
Example 1:
Input: [1,2,3,3] Output: 3
Example 2:
Input: [2,1,2,5,3,2] Output: 2
Example 3:
Input: [5,1,5,2,5,3,5,4] Output: 5
Note:
4 <= A.length <= 10000
0 <= A[i] < 10000
A.length
is even
class Solution:
def repeatedNTimes(self, nums: List[int]) -> int:
s = set()
for num in nums:
if num in s:
return num
s.add(num)
class Solution {
public int repeatedNTimes(int[] nums) {
Set<Integer> s = new HashSet<>();
for (int num : nums) {
if (s.contains(num)) {
return num;
}
s.add(num);
}
return -1;
}
}
class Solution {
public:
int repeatedNTimes(vector<int>& nums) {
unordered_set<int> s;
for (auto &num : nums) {
if (s.find(num) != s.end()) {
return num;
}
s.insert(num);
}
return -1;
}
};
/**
* @param {number[]} nums
* @return {number}
*/
var repeatedNTimes = function (nums) {
const s = new Set();
for (const num of nums) {
if (s.has(num)) {
return num;
}
s.add(num);
}
return -1;
};