This was one of the problems in LeetCode's ongoing 30-day challenge. Published on 7th April 2020.
Given an integer array arr, count element x such that x + 1 is also in arr.
If there're duplicates in arr, count them separately.
Input: arr = [1,2,3] Output: 2 Explanation: 1 and 2 are counted cause 2 and 3 are in arr.
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.
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.
Input: arr = [1,1,2,2] Output: 2 Explanation: Two 1s are counted cause 2 is in arr.
- As the original list contains duplicate elements as well. Convert the list to set to get unique elements.
- Iterate over the set and check if element+1 exists in the original list.
- If it exists. Find the count of the element in the original list.
- Keep adding the count to the new list.
- Sum of all the counts. Returns the required output.
class Solution: def countElements(self, arr: List[int]) -> int: unique_elements = set(arr) L =  for element in unique_elements: if element + 1 in arr: item_count = arr.count(element) L.append(item_count) return sum(L)
Discussion on more optimized solutions are welcome.
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