[LeetCode] (#) 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

class Solution {
    public int countElements(int[] arr) {
        int i, j, cnt = 0, n = arr.length;
        int[] tmp = new int[n];
        for( i = 0; i < n; i++ )
        	tmp[i] = arr[i]+1;
        loop:
	for( i = 0; i < n; i++ ) {
        	for( j = 0; j < n; j++ ) {
        		if( tmp[i] == arr[j] ) {
        			cnt++;
        			continue loop;
        		}
        	}
       	}
	return cnt;
    }
}