1432-max-difference-you-can-get-from-changing-an-integer.py

"""
Problem Link: https://leetcode.com/problems/max-difference-you-can-get-from-changing-an-integer/

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

Example 3:
Input: num = 123456
Output: 820000

Example 4:
Input: num = 10000
Output: 80000

Example 5:
Input: num = 9288
Output: 8700
 
Constraints:
1 <= num <= 10^8
"""
class Solution:
    def maxDiff(self, num: int) -> int:
        num = list(str(num))
        first_digit = num[0]
        x = '9'
        y = num[0] if num[0] != '1' else '0'
        
        for i in num:
          if i != x:
            x = i
            break
        
        if y == '0':
          for i in range(1, len(num)):
            if num[i] not in ['0','1']:
              y = num[i]
              break
        
        num1 = [i for i in num]
        for i in range(len(num)):
          if num[i] == x:
            num[i] = '9'
          if num1[i] == y:
            num1[i] = '1' if first_digit != '1' else '0'
            
        
        return int("".join(num)) - int("".join(num1))