Counting Elements Python

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

 Python : Solution 1

class Solution:

    def countElements(self, arr: List[int]) -> int:

        count = 0

        for x in arr:

            if x+1 in arr:

                count+=1

        return count

Time Complexity : O(n*n) because search in entire list

Space Complexity : O(1)

Python : Solution 2

class Solution:

    def countElements(self, arr: List[int]) -> int:

        hash_set = set(arr)

        count = 0

        for x in arr:

            if x+1 in hash_set:

                count+=1

        return count

Time Complexity : O(n) because search in hash set is O(1) 

Space Complexity : O(n)

Python Solution 3

class Solution:

    def countElements(self, arr: List[int]) -> int:

        arr.sort()

        count = 0

        run_length = 1

        for i in range(len(arr)):

            if arr[i-1] != arr[i]:

                if arr[i-1]+1 == arr[i]:

                    count +=run_length

                run_length = 0

            run_length+=1

        return count

Time Complexity : O(nlogn)

Space Complexity : O(n) to O(1)