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Count Number of Distinct Integers After Reverse Operations

Difficulty: Medium

Problem Description

You are given an array nums consisting of positive integers. You have to take each integer in the array, reverse its digits, and add it to the end of the array. You should apply this operation to the original integers in nums. Return the number of distinct integers in the final array.

Key Insights

  • Each integer can be reversed, which may lead to duplicates.
  • The reversed integer for numbers like 10 and 01 will result in the same integer.
  • We need to count only distinct integers.

Space and Time Complexity

Time Complexity: O(n), where n is the number of elements in the input array nums. Each number in the array is processed in constant time. Space Complexity: O(n), for storing distinct integers in a set.

Solution

To solve the problem, we can use a set to store distinct integers. For each integer in the input array, we will reverse its digits, convert it back to an integer, and add both the original and the reversed integers to the set. Finally, the size of the set will give us the count of distinct integers.

Code Solutions

def count_distinct_integers(nums):
    distinct_integers = set()
    
    for num in nums:
        # Add the original number
        distinct_integers.add(num)
        
        # Reverse the digits and convert back to an integer
        reversed_num = int(str(num)[::-1])
        distinct_integers.add(reversed_num)
    
    return len(distinct_integers)
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