Introduction

In this blog post, we will explore how to solve the LeetCode problem "2566" 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 integer num. You know that Bob will sneakily remap one of the 10 possible digits (0 to 9) to another digit. Return the difference between the maximum and minimum values Bob can make by remapping exactly one digit in num. Notes: When Bob remaps a digit d1 to another digit d2, Bob replaces all occurrences of d1 in num with d2. Bob can remap a digit to itself, in which case num does not change. Bob can remap different digits for obtaining minimum and maximum values respectively. The resulting number after remapping can contain leading zeroes. Example 1: Input: num = 11891 Output: 99009 Explanation: To achieve the maximum value, Bob can remap the digit 1 to the digit 9 to yield 99899. To achieve the minimum value, Bob can remap the digit 1 to the digit 0, yielding 890. The difference between these two numbers is 99009. Example 2: Input: num = 90 Output: 99 Explanation: The maximum value that can be returned by the function is 99 (if 0 is replaced by 9) and the minimum value that can be returned by the function is 0 (if 9 is replaced by 0). Thus, we return 99. Constraints: 1 <= num <= 108

Explanation

Here's the solution to the problem:

• Find Remapping for Maximum: Iterate through the digits of the number. The first digit that is not '9' is the digit to remap to '9'.
• Find Remapping for Minimum: Iterate through the digits of the number. The first digit that is not '0' (and also not the first digit if the first digit is '1') is the digit to remap to '0'. If the first digit is '1', look for any other digit that is not zero, and remap that to 0. If all other digits are zero, remap '1' to '0'.

• Calculate Difference: Compute the maximum and minimum numbers after remapping, and return their difference.

• Runtime Complexity: O(N), where N is the number of digits in num.

• Storage Complexity: O(N)

Code

    def solve():
    num = int(input())
    s_num = str(num)

    def get_max(s):
        for digit in s:
            if digit != '9':
                replace_digit = digit
                break
        else:
            return int(s)

        new_s = s.replace(replace_digit, '9')
        return int(new_s)

    def get_min(s):
        first_digit = s[0]

        if first_digit != '1':
            for digit in s:
                if digit != '0':
                    replace_digit = digit
                    break
            else:
                return int(s)
        else:
            if len(s) > 1:
                for digit_index in range(1, len(s)):
                    if s[digit_index] != '0':
                        replace_digit = s[digit_index]
                        break
                else:
                    replace_digit = first_digit # all other digits are zero, change the first digit
            else:
                return 0

        new_s = s.replace(replace_digit, '0')
        return int(new_s)

    maximum = get_max(s_num)
    minimum = get_min(s_num)
    return maximum - minimum


print(solve())