ALGORITHM CHALLENGE: 2

Algorithms and Challenge Questions — Sat, 22 Jun 2019 15:56:55 +0000

Challenge

Return the number (count) of vowels in the given string.
We will consider a, e, i, o, and u as vowels for this Kata. The input string will only consist of lower case letters and/or spaces.

Solution

rust

fn get_count(string: &str) -> usize {
  let vowels: &str = "aeiou";
  let mut vowels_count: usize = 0;
  for c in string.chars() {
      if vowels.contains(c) {
          vowels_count = vowels_count + 1; 
      }
  }
  vowels_count
}

ALGORITHM CHALLENGE: 1

Algorithms and Challenge Questions — Wed, 19 Jun 2019 06:26:44 +0000

Challenge

This time no story, no theory. The examples below show you how to write function accum:

Examples:

accum("abcd") -> "A-Bb-Ccc-Dddd"
accum("RqaEzty") -> "R-Qq-Aaa-Eeee-Zzzzz-Tttttt-Yyyyyyy"
accum("cwAt") -> "C-Ww-Aaa-Tttt"

The parameter of accum is a string which includes only letters from a..z and A..Z.

Solution

fn format_string(word: &str) -> String{
    let mut result: Vec<char> = vec![] ; //This is where we store our result which is array of character
    let mut conversion: Vec<char>; 
    for (i,c) in  word.chars().enumerate(){
         if i > 0{
            result.append(&mut vec!['-']);
         }
         result.append(&mut c.to_uppercase().collect());
         conversion = c.to_lowercase().collect();
         for _x in 0..i {
            // you need to clone else it prints seems a series of same object append and print only one
            result.append(&mut conversion.clone());
         }
    }
    return result.iter().collect();
}
fn main(){
     println!("{:?}", format_string("RqaEzty"))
}

//go implementation coming soon

DEV Community: Algorithms and Challenge Questions

ALGORITHM CHALLENGE: 2

Challenge

Solution

rust

ALGORITHM CHALLENGE: 1

Challenge

Solution