Introduction

In this blog post, we will explore how to solve the LeetCode problem "1432" 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 will apply the following steps exactly two times: Pick a digit x (0 <= x <= 9). Pick another digit y (0 <= y <= 9). The digit y can be equal to x. Replace all the occurrences of x in the decimal representation of num by y. The new integer cannot have any leading zeros, also the new integer cannot be 0. Let a and b be the results of applying the operations to num the first and second times, respectively. Return the max difference between a and b. Example 1: Input: num = 555 Output: 888 Explanation: The first time pick x = 5 and y = 9 and store the new integer in a. The second time pick x = 5 and y = 1 and store the new integer in b. We have now a = 999 and b = 111 and max difference = 888 Example 2: Input: num = 9 Output: 8 Explanation: The first time pick x = 9 and y = 9 and store the new integer in a. The second time pick x = 9 and y = 1 and store the new integer in b. We have now a = 9 and b = 1 and max difference = 8 Constraints: 1 <= num <= 108

Explanation

• Maximize a and Minimize b: To maximize the difference a - b, we need to find the largest possible value for a and the smallest possible value for b after performing the two replacement operations.
  • Maximize by Replacing with 9: For maximizing a, we replace the first digit we encounter with 9, unless the number already starts with 9. If the first digit is not 9, we replace it with 9. Otherwise, we iterate through the digits to find a digit that is not 9.
  • Minimize by Replacing with 0 or 1: For minimizing b, if the first digit is 1, we want to find some other digit that is not 0 or 1 and replace it with 0. If the first digit is not 1, replace the first digit with 1.

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

Code

    def maxDiff(num: int) -> int:
    s = str(num)
    n = len(s)

    def replace(s, x, y):
        new_s = s.replace(x, y)
        if new_s[0] == '0' and len(new_s) > 1:
            return -1
        return int(new_s) if new_s != "0" else -1

    a = num
    b = num

    # Maximize a
    for i in range(n):
        if s[i] != '9':
            temp_a = replace(s, s[i], '9')
            if temp_a != -1:
              a = temp_a
            break

    # Minimize b
    if s[0] != '1':
      temp_b = replace(s, s[0], '1')
      if temp_b != -1:
        b = temp_b

    else:
      for i in range(1, n):
        if s[i] != '0' and s[i] != '1':
          temp_b = replace(s, s[i], '0')
          if temp_b != -1:
            b = temp_b
          break


    return a - b