Introduction

In this blog post, we will explore how to solve the LeetCode problem "1505" 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 string num representing the digits of a very large integer and an integer k. You are allowed to swap any two adjacent digits of the integer at most k times. Return the minimum integer you can obtain also as a string. Example 1: Input: num = "4321", k = 4 Output: "1342" Explanation: The steps to obtain the minimum integer from 4321 with 4 adjacent swaps are shown. Example 2: Input: num = "100", k = 1 Output: "010" Explanation: It's ok for the output to have leading zeros, but the input is guaranteed not to have any leading zeros. Example 3: Input: num = "36789", k = 1000 Output: "36789" Explanation: We can keep the number without any swaps. Constraints: 1 <= num.length <= 3 * 104 num consists of only digits and does not contain leading zeros. 1 <= k <= 109

Explanation

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

• High-Level Approach:

  • Iterate through the input num string from left to right.
  • For each position, find the smallest digit to the right that can be moved to the current position within the allowed k swaps.
  • Update k (the remaining swaps) and the num string accordingly by performing the swap.

• Complexity:

  • Runtime Complexity: O(n^2), where n is the length of the input string num.
  • Storage Complexity: O(n), primarily for storing the modified string as a list.

Code

    def minInteger(num: str, k: int) -> str:
    n = len(num)
    num_list = list(num)

    for i in range(n):
        if k == 0:
            break

        min_digit = num_list[i]
        min_index = i

        for j in range(i + 1, min(n, i + k + 1)):
            if num_list[j] < min_digit:
                min_digit = num_list[j]
                min_index = j

        if min_index != i:
            for j in range(min_index, i, -1):
                num_list[j], num_list[j - 1] = num_list[j - 1], num_list[j]
            k -= (min_index - i)

    return "".join(num_list)