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3289. The Two Sneaky Numbers of Digitville

Description

In the town of Digitville, there was a list of numbers called nums containing integers from 0 to n - 1. Each number was supposed to appear exactly once in the list, however, two mischievous numbers sneaked in an additional time, making the list longer than usual.

As the town detective, your task is to find these two sneaky numbers. Return an array of size two containing the two numbers (in any order), so peace can return to Digitville.

 

Example 1:

Input: nums = [0,1,1,0]

Output: [0,1]

Explanation:

The numbers 0 and 1 each appear twice in the array.

Example 2:

Input: nums = [0,3,2,1,3,2]

Output: [2,3]

Explanation:

The numbers 2 and 3 each appear twice in the array.

Example 3:

Input: nums = [7,1,5,4,3,4,6,0,9,5,8,2]

Output: [4,5]

Explanation:

The numbers 4 and 5 each appear twice in the array.

 

Constraints:

• 2 <= n <= 100
• nums.length == n + 2
• 0 <= nums[i] < n
• The input is generated such that nums contains exactly two repeated elements.

Solutions

Solution 1: Counting

We can use an array $\textit{cnt}$ to record the number of occurrences of each number.

Traverse the array $\textit{nums}$, and when a number appears for the second time, add it to the answer array.

After the traversal, return the answer array.

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

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

• class Solution {
      public int[] getSneakyNumbers(int[] nums) {
          int[] ans = new int[2];
          int[] cnt = new int[100];
          int k = 0;
          for (int x : nums) {
              if (++cnt[x] == 2) {
                  ans[k++] = x;
              }
          }
          return ans;
      }
  }
  
  

• class Solution {
  public:
      vector<int> getSneakyNumbers(vector<int>& nums) {
          vector<int> ans;
          int cnt[100]{};
          for (int x : nums) {
              if (++cnt[x] == 2) {
                  ans.push_back(x);
              }
          }
          return ans;
      }
  };
  
  

• class Solution:
      def getSneakyNumbers(self, nums: List[int]) -> List[int]:
          cnt = Counter(nums)
          return [x for x, v in cnt.items() if v == 2]
  
  

• func getSneakyNumbers(nums []int) (ans []int) {
  	cnt := [100]int{}
  	for _, x := range nums {
  		cnt[x]++
  		if cnt[x] == 2 {
  			ans = append(ans, x)
  		}
  	}
  	return
  }
  
  

• function getSneakyNumbers(nums: number[]): number[] {
      const ans: number[] = [];
      const cnt: number[] = Array(100).fill(0);
      for (const x of nums) {
          if (++cnt[x] > 1) {
              ans.push(x);
          }
      }
      return ans;
  }
  
  

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