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  1. Separate the Numbers
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Separate the Numbers

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101 Discussions

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  • josh_b_merrell
     + 0 comments

    I spent like 10 hours trying to do this the hard way. When I figured out the easy way I did it in 5 minutes during a business meeting. Do yourself a favor:

    For input "99100", grab the first digit 9 and build beautiful string 91011, and check whether it's identical to 99100, It's not. So grab the first 2 digits 99 and build beautiful string 99100. It's identical, so "YES 99".

    Don't try to stairstep a left index and a right index or dynamically keep track of curernt index length or anything. Save it for the harder questions lmao.

  • kocsis_tamas_ol1
     + 0 comments

    I have no clue in whos vocab is this a Basic - Easy questions. This is more like a medium question and I would say a harder one of that.

    I have used a backtracking kind approach. Try diff length numbers to begin the recursive calls, then check if the next number we can come up with is +1 of previous. If we find a number like that we have another recursive call.| If the number is too big we can stop and return false. If the starting number size is already more than half the digits count we can also stop.

  • eliszus
     + 1 comment

    Let's talk about constraints...

    The constraints listed on this problem are EXTREMELY MISLEADING! There was a crazy amount of guessing on this problem to figure out which constraints the hackerrank brain was assuming since the ones written did not work.

  • jonas_vdheyden
     + 0 comments
    def separateNumbers(s):
    # Write your code here
    i=1
    while i-1 < len(s)//2:
        d=int(s[:i])
        if rec(d,s[i:]):
            print("YES "+s[:i])
            return
        else:
            i+=1
    print("NO")
    return
            
def rec(d,s):
    l = len(str(d+1))
    if len(s) < l:
        return False
    d_ = int(s[:l])
    if d_ == d+1:
        if len(s[l:])>0:
            return rec(d_,s[l:])
        else:
            return True
    else:
        return False

  • victor_reial
     + 0 comments

    Rust best solution

    If you’re looking for solutions to the 3-month preparation kit in either Python or Rust, you can find them below: my solutions 

    pub fn separate_numbers(s: &str) {
    //Time complexity: O(n^2)
    //Space complexity (ignoring input): O(n)
    //Python solution is way cleaner, made this one first and got lazy to rewrite
    let len_s = s.len();
    if len_s == 1 {
        println!("NO");
        return;
    }
    let vec_chars = s.chars().collect::<Vec<char>>();
    let mut start_number_size = 1;
    let mut number_size = start_number_size;
    let mut index: usize = start_number_size;
    let mut last_number = vec_chars[0]
        .to_string()
        .parse::<i64>()
        .expect("To have a number");
    let mut first_number = last_number;
    loop {
        if (last_number + 1) == (10u64.pow(number_size as u32) as i64) {
            number_size += 1;
        }
        let number = if number_size + index > len_s {
            last_number + 2
        } else {
            vec_chars[index..(number_size + index)]
                .iter()
                .collect::<String>()
                .parse::<i64>()
                .expect("To have a number")
        };
        if ((number - last_number) != 1) || (index + number_size > len_s) {
            start_number_size += 1;
            if start_number_size > (len_s + 1) / 2 {
                println!("NO");
                return;
            };
            number_size = start_number_size;
            index = 0;
            if vec_chars[0] == '0' {
                println!("NO");
                return;
            }
            last_number = vec_chars[0..start_number_size]
                .iter()
                .collect::<String>()
                .parse::<i64>()
                .expect("To have a number");
            first_number = last_number;
        } else {
            last_number = number;
        }
        index += number_size;
        if index == len_s {
            println!("YES {first_number}");
            return;
        }
    }
}