Formatted question description: https://leetcode.ca/all/1426.html
1426. Counting Elements
Level
Easy
Description
Given an integer array arr, count how many elements x there are, 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.
Example 5:
Input: arr = [1,1,2]
Output: 2
Explanation: Both 1s are counted because 2 is in the array.
Constraints:
1 <= arr.length <= 10000 <= arr[i] <= 1000
Solution
Use a set to store all the elements in arr. Then loop over arr. For each element x in arr, if x + 1 is in the set, then increase the counter by 1. Finally, return the counter.
class Solution {
public int countElements(int[] arr) {
Set<Integer> set = new HashSet<Integer>();
for (int num : arr)
set.add(num);
int count = 0;
for (int num : arr) {
if (set.contains(num + 1))
count++;
}
return count;
}
}