Introduction

In this blog post, we will explore how to solve the LeetCode problem "2442" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. Example 1: Input: nums = [1,13,10,12,31] Output: 6 Explanation: After including the reverse of each number, the resulting array is [1,13,10,12,31,1,31,1,21,13]. The reversed integers that were added to the end of the array are underlined. Note that for the integer 10, after reversing it, it becomes 01 which is just 1. The number of distinct integers in this array is 6 (The numbers 1, 10, 12, 13, 21, and 31). Example 2: Input: nums = [2,2,2] Output: 1 Explanation: After including the reverse of each number, the resulting array is [2,2,2,2,2,2]. The number of distinct integers in this array is 1 (The number 2). Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 106

Explanation

• Core Idea: Iterate through the input array nums. For each number, reverse its digits and add the reversed number to a set. The set efficiently stores only unique numbers, preventing duplicates. Finally, return the size of the set.

• Optimization: Use a set data structure to automatically handle duplicate values and ensure we only count distinct integers. The digit reversal is done efficiently via string slicing.

• Complexity: O(N * log(M)) runtime, where N is the length of nums and M is the maximum value in nums. This is due to the iteration through nums and the log(M) factor coming from the string conversion and reversal for numbers up to M. The space complexity is O(N) in the worst case, to store the unique numbers in the seen set.

Code

    def countDistinctIntegers(nums):
    seen = set(nums)
    for num in nums:
        reversed_num = int(str(num)[::-1])
        seen.add(reversed_num)
    return len(seen)