Introduction

In this blog post, we will explore how to solve the LeetCode problem "2343" 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 a 0-indexed array of strings nums, where each string is of equal length and consists of only digits. You are also given a 0-indexed 2D integer array queries where queries[i] = [ki, trimi]. For each queries[i], you need to: Trim each number in nums to its rightmost trimi digits. Determine the index of the kith smallest trimmed number in nums. If two trimmed numbers are equal, the number with the lower index is considered to be smaller. Reset each number in nums to its original length. Return an array answer of the same length as queries, where answer[i] is the answer to the ith query. Note: To trim to the rightmost x digits means to keep removing the leftmost digit, until only x digits remain. Strings in nums may contain leading zeros. Example 1: Input: nums = ["102","473","251","814"], queries = [[1,1],[2,3],[4,2],[1,2]] Output: [2,2,1,0] Explanation: 1. After trimming to the last digit, nums = ["2","3","1","4"]. The smallest number is 1 at index 2. 2. Trimmed to the last 3 digits, nums is unchanged. The 2nd smallest number is 251 at index 2. 3. Trimmed to the last 2 digits, nums = ["02","73","51","14"]. The 4th smallest number is 73. 4. Trimmed to the last 2 digits, the smallest number is 2 at index 0. Note that the trimmed number "02" is evaluated as 2. Example 2: Input: nums = ["24","37","96","04"], queries = [[2,1],[2,2]] Output: [3,0] Explanation: 1. Trimmed to the last digit, nums = ["4","7","6","4"]. The 2nd smallest number is 4 at index 3. There are two occurrences of 4, but the one at index 0 is considered smaller than the one at index 3. 2. Trimmed to the last 2 digits, nums is unchanged. The 2nd smallest number is 24. Constraints: 1 <= nums.length <= 100 1 <= nums[i].length <= 100 nums[i] consists of only digits. All nums[i].length are equal. 1 <= queries.length <= 100 queries[i].length == 2 1 <= ki <= nums.length 1 <= trimi <= nums[i].length Follow up: Could you use the Radix Sort Algorithm to solve this problem? What will be the complexity of that solution?

Explanation

Here's a breakdown of the approach and the Python code:

• High-Level Approach:

  • For each query, extract the k and trim values.
  • Create a list of tuples where each tuple contains the trimmed string and its original index.
  • Sort the list of tuples based on the trimmed string (lexicographically). The original index is used as a tie-breaker.
  • Return the index of the k-th smallest element from the sorted list.

• Complexity:

  • Time Complexity: O(q * (n log n)), where q is the number of queries and n is the length of nums. The n log n part comes from sorting the trimmed strings for each query.
  • Space Complexity: O(n) due to the creation of the temporary list of tuples for each query.

Code

    def smallestTrimmedNumbers(nums: list[str], queries: list[list[int]]) -> list[int]:
    """
    Finds the index of the k-th smallest trimmed number in nums for each query.

    Args:
        nums: A list of strings, where each string is of equal length and consists of only digits.
        queries: A 2D integer array where queries[i] = [ki, trimi].

    Returns:
        An array answer of the same length as queries, where answer[i] is the answer to the ith query.
    """

    answer = []
    for k, trim in queries:
        trimmed_nums = []
        for i, num in enumerate(nums):
            trimmed_nums.append((num[-trim:], i))  # Trim and store index

        trimmed_nums.sort()  # Sort based on trimmed string and index

        answer.append(trimmed_nums[k - 1][1])  # Append the index of the k-th smallest

    return answer