LeetCode - Counting Elements - 30Days Challenge

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

We can use HashSet to do fast lookup for all a+1 elements.

Code


import java.util.HashSet;
import java.util.Set;

public class CountingElements {

	public int countElements(int[] arr) {

		Set<Integer> set = new HashSet<Integer>();

		for (int a : arr) {
			set.add(a);
		}

		int cnt = 0;

		for (int a : arr) {
			if (set.contains(a + 1))
				cnt++;
		}

		return cnt;

	}
	

	public static void main(String[] args) {
		int [] a= {1,2,3};
		System.out.println(new CountingElements().countElements(a));
	}
}