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Solution.java
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Solution.java
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package countingelements;
import java.util.HashSet;
class Solution {
/**
* Given an integer array arr, count element x such that x + 1 is also in arr.
* <p>
* If there're duplicates in arr, count them seperately.
* <p>
* <p>
* <p>
* Example 1:
* <p>
* Input: arr = [1,2,3]
* Output: 2
* Explanation: 1 and 2 are counted cause 2 and 3 are in arr.
* Example 2:
* <p>
* 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:
* <p>
* 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:
* <p>
* Input: arr = [1,1,2,2]
* Output: 2
* Explanation: Two 1s are counted cause 2 is in arr.
* <p>
* <p>
* Constraints:
* <p>
* 1 <= arr.length <= 1000
* 0 <= arr[i] <= 1000
*/
int countElements(int[] arr) {
HashSet<Integer> nums = new HashSet<>();
for (int i : arr)
nums.add(i);
int count = 0;
for (int x : arr) {
if (nums.contains(x + 1))
count++;
}
return count;
}
/*
35 / 35 test cases passed.
Status: Accepted
Runtime: 3 ms
Memory Usage: 39.6 MB
*/
}