leetcode 30 day challenge Counting Elements

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

Ｓｏｌｕｔｉｏｎ

思路：用一个集合存储元素，再遍历即可

class Solution:
    def countElements(self, arr: List[int]) -> int:
        myset = set(arr)
        count = 0
        for i in arr:
            if i+1 in myset:
                count += 1
        return count

分析：
时间复杂度：O(N)
空间复杂度：O(N)