LeetCode 30 Day Challenge | Day 7 | Counting Elements

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

C++ Solution:

class Solution {
public:
    int countElements(vector<int>& arr) {
        unordered_map<int, int> _map;
        
        for(int a : arr){
            if(_map.find(a) != _map.end())
                _map[a]++;
            else
                _map[a] = 1;
        }
        
        int result = 0;
        for(auto n : _map){
            if(_map.find(n.first + 1) != _map.end())
                result += n.second;
        }
        
        return result;
    }
};

Detailed Explanation on Youtube: