[LeetCode] Weekly 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

这道题是时间和空间的复杂度都是O(N), 想法比较直接，丢到map里，然后loop map，如果k+1也在map里，就把k的val加到count上。注意审题，算count的时候是算k的val

class Solution:
    def countElements(self, arr: List[int]) -> int:
        if not arr: return 0
        
        d = collections.defaultdict(int)
        for num in arr:
            d[num] += 1
            
        count = 0
        for k in d:
            if k+1 in d:
                count += d[k]
        return count