Welcome to Subscribe On Youtube

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

Solution 1: Hash Table

Since the array $\textit{nums}$ has a total of $2n$ elements, with $n + 1$ distinct elements, and one element repeated $n$ times, this means the remaining $n$ elements in the array are all distinct.

Therefore, we only need to iterate through the array $\textit{nums}$ and use a hash table $s$ to record the elements we’ve encountered. When we encounter an element $x$, if $x$ already exists in the hash table $s$, it means $x$ is the repeated element, and we can directly return $x$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $\textit{nums}$.

Solution 2: Mathematics

According to the problem description, half of the elements in the array $\textit{nums}$ are the same. If we view the array as a circular arrangement, then there is at most $1$ other element between two identical elements.

Therefore, we iterate through the array $\textit{nums}$ starting from index $2$. For each index $i$, we compare $\textit{nums}[i]$ with $\textit{nums}[i - 1]$ and $\textit{nums}[i - 2]$. If they are equal, we return that value.

If we don’t find the repeated element in the above process, then the repeated element must be $\textit{nums}[0]$, and we can directly return $\textit{nums}[0]$.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• 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];
              }
          }
      }
  }
  
  
  // Solution 2
  class Solution {
      public int repeatedNTimes(int[] nums) {
          for (int i = 2; i < nums.length; ++i) {
              if (nums[i] == nums[i - 1] || nums[i] == nums[i - 2]) {
                  return nums[i];
              }
          }
          return nums[0];
      }
  }
  
  

• 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]);
          }
      }
  };
  
  
  // Solution 2
  class Solution {
  public:
      int repeatedNTimes(vector<int>& nums) {
          for (int i = 2; i < nums.size(); ++i) {
              if (nums[i] == nums[i - 1] || nums[i] == nums[i - 2]) {
                  return nums[i];
              }
          }
          return nums[0];
      }
  };
  
  

• class Solution:
      def repeatedNTimes(self, nums: List[int]) -> int:
          s = set()
          for x in nums:
              if x in s:
                  return x
              s.add(x)
  
  
  # Solution 2
  class Solution:
      def repeatedNTimes(self, nums: List[int]) -> int:
          for i in range(2, len(nums)):
              if nums[i] == nums[i - 1] or nums[i] == nums[i - 2]:
                  return nums[i]
          return nums[0]
  
  

• func repeatedNTimes(nums []int) int {
  	s := map[int]bool{}
  	for i := 0; ; i++ {
  		if s[nums[i]] {
  			return nums[i]
  		}
  		s[nums[i]] = true
  	}
  }
  
  
  // Solution 2
  func repeatedNTimes(nums []int) int {
  	for i := 2; i < len(nums); i++ {
  		if nums[i] == nums[i-1] || nums[i] == nums[i-2] {
  			return nums[i]
  		}
  	}
  	return nums[0]
  }
  
  

• 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);
      }
  }
  
  
  // Solution 2
  function repeatedNTimes(nums: number[]): number {
      for (let i = 2; i < nums.length; ++i) {
          if (nums[i] === nums[i - 1] || nums[i] === nums[i - 2]) {
              return nums[i];
          }
      }
      return nums[0];
  }
  
  

• /**
   * @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);
      }
  };
  
  
  // Solution 2
  /**
   * @param {number[]} nums
   * @return {number}
   */
  var repeatedNTimes = function (nums) {
      for (let i = 2; i < nums.length; ++i) {
          if (nums[i] === nums[i - 1] || nums[i] === nums[i - 2]) {
              return nums[i];
          }
      }
      return nums[0];
  };
  
  

All Problems

All Solutions