README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0900-0999/0961.N-Repeated%20Element%20in%20Size%202N%20Array/README_EN.md	Array Hash Table

961. N-Repeated Element in Size 2N Array

中文文档

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

Solution 1

Python3

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

Java

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

C++

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]);
        }
    }
};

Go

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

TypeScript

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

JavaScript

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