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 <= 10000 <= 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:
