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counting-elements

Problem

Problem Description

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

Counting Elemnents
Time complexity:
O(N)
N - the length of array length
class Solution {
public int countElements(int[] arr) {
if (arr == null || arr.length < 2) return 0;
int count = 0;
Set<Integer> set = new HashSet<>();
for (int num : arr) {
set.add(num);
}
for (int num : arr) {
count += set.contains(num + 1) ? 1 : 0;
}
return count;
}
}