Naive solution to problem

Char Mapping.rs

/*

    Determine whether there exists a one-to-one character mapping from one string s1 to another s2.
    
    For example, given s1 = abc and s2 = bcd, return true since we can map a to b, b to c, and c to d.
    
    Given s1 = foo and s2 = bar, return false since the o cannot map to two characters.

*/

use super::Problem;

pub struct Problem578 {

}

impl Problem578 {

    pub fn new() -> Self {
        Problem578 {}
    }

    pub fn can_map(&self, str1: &str, str2: &str) -> bool {

        if str1.len() != str2.len() {
            return false
        }

        let chars1 = str1.as_bytes();
        let chars2 = str2.as_bytes();

        let len = str1.len();

        let mut sum1 = 0u64;
        let mut sum2 = 0u64;

        for i in 0..len {

            let b1 = (chars1[i] as u64) << 1;
            let b2 = (chars2[i] as u64) << 1;

            match (sum1).checked_add(b1 as u64) {
                None => return false,
                Some(n) => sum1 = n
            }

            match (sum2).checked_add(b2 as u64) {
                None => return false,
                Some(n) => sum2 = n
            }
        }

        println!("sum1: {}, sum2: {}", sum1, sum2);

        return sum1 == sum2;
    }
}

impl Problem for Problem578 {

    fn run(&self) {
        let str1 = String::from("abcdefghijklmnopqrstuvwxyz");
        let str2 = String::from("zyxwvutsrqpomnlkjihgfedcba");
        
        let result = self.can_map(str1.as_str(), str2.as_str());

        println!("Result: {}", result);
    }

}