Given an integer array arr, count element x such that x + 1 is also in arr.

If there’re duplicates in arr, count them seperately.

 

Example 1:

Input: arr = [1,2,3] Output: 2 Explanation: 1 and 2 are counted cause 2 and 3 are in arr.

Example 2:

Input: arr = [1,1,3,3,5,5,7,7] Output: 0 Explanation: No numbers are counted, cause there’s no 2, 4, 6, or 8 in arr.

Example 3:

Input: arr = [1,3,2,3,5,0] Output: 3 Explanation: 0, 1 and 2 are counted cause 1, 2 and 3 are in arr.

Example 4:

Input: arr = [1,1,2,2] Output: 2 Explanation: Two 1s are counted cause 2 is in arr.

 

Constraints:

  • 1 <= arr.length <= 1000
  • 0 <= arr[i] <= 1000

Solution

We can use HashSet to do fast lookup for all a+1 elements.

Code

Output

2

We encourage you to write a comment if you have a better solution or having any doubt on the above topic.

Leave a Reply

Your email address will not be published. Required fields are marked *