Counting Elements Problem statement:

LeetCode

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

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

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

Counting Elements ES6 solution

/**
 * @param {number[]} arr
 * @return {number}
 */
var countElements = function(arr) {
    let count = 0;
    for(let i = 0; i < arr.length; i++){
        if(arr.includes(arr[i]+1))
            count++;
        } 
    }
    return count;
};

Submission Output on LeetCode: