README.md

1432. 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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn max_diff(num: i32) -> i32 {
        let num = num.to_string();
        let max = match num.chars().find(|&c| c < '9') {
            Some(c) => num.replace(c, "9"),
            None => num.clone(),
        };
        let min = if num.starts_with('1') {
            match num.chars().find(|&c| c > '1') {
                Some(c) => num.replace(c, "0"),
                None => num,
            }
        } else {
            num.replace(num.chars().next().unwrap(), "1")
        };

        max.parse::<i32>().unwrap() - min.parse::<i32>().unwrap()
    }
}