Soyo

Day 7 of 30 days Leetcode challenge

Counting Element
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

Solutions

Use hashtable to search existence of element + 1, take O(2n) time.

class Solution {
    func countElements(_ arr: [Int]) -> Int {
        var hash = Set(arr)
        var res = 0
        for ele in arr {
            if hash.contains(ele + 1) {
                res += 1
            }
        }
        return res
    }
}

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