Leetcode: 2442. Count Number of Distinct Integers After Reverse Operations
Understand the problem
Given an array \(a\), find the number of distinct elements on \(b\) where \(b\) contains elements of \(a\) and for each \(x\) in \(a\), it also contains the number generated by reversing the digits of \(x\).
Useful prompts
Devise a plan
Pattern: Small constraints could allow a brute-force solution, so generate \(b\) and return its size. Time complexity and space are \(O(n)\).
Carry out the plan
class Solution: def countDistinctIntegers(self, nums: List[int]) -> int: seen = set(nums) for n in nums: seen.add(int(str(n)[::-1])) return len(seen)