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# 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

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;

}

}

Last modified 3yr ago