index.js

// 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

const countElements = (arr) => {
    const mySet = new Set();

    for (let val in arr) {
        mySet.add(arr[+val])
    }

    let res = 0;

    for (let key of arr) {
        if (mySet.has(+key + 1)) {
            res += 1
        }
    }
    return res
};