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
Hide Hint #1
Use hashset to store all elements. Hide Hint #2
Loop again to count all valid elements.
class Solution:
def countElements(self, arr: List[int]) -> int:
return len([n for n in arr if n + 1 in set(arr)])