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  1. Prepare
  2. Algorithms
  3. Strings
  4. Super Reduced String
  5. Discussions

Super Reduced String

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1725 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/XgJKCkb1EjQ

    #include <bits/stdc++.h>

using namespace std;

int main()
{
    string s;
    cin >> s;
    int i = 1;
    while(i < s.size()){
        if(s[i-1] == s[i]){
            s.erase(i-1, 2);
            if( i != 1) i--;
        }
        else i++;
    }
    if(s == "") s = "Empty String";
    cout << s;
    return 0;
}

  • sayeedahmad06061
     + 0 comments

    def superReducedString(s): st=0 end=1 for i in range(len(s)): try: if(s[st]==s[end]): s=s[:st]+s[end+1:] st -= 1 end -= 1 st=max(st,0) end=max(end,1) print(s,st,end) else: st += 1 end += 1 except IndexError: return s if s else 'Empty String' return s

  • ddebajyati
     + 0 comments

    Best and performant approach in C++. Behold my simple C++ solution -

    string superReducedString(string s) {
    int i=1;
    while (i<s.length()) {
        if (s[i] == s[i-1]) {
            s.erase(i-1,2);
            i=1;
        } else {
            i++;
        }
    }
    
    if (s.empty()) {
        return "Empty String";
    } else {
        return s;
    }
}

  • kpyadav_bbsr
     + 0 comments

    python recursive implementation

    def superReducedString(s):
    # Write your code here
    
    if len(s) == 0:
        return 'Empty String'
    elif len(s) == 1:
        return s
    elif all([s[i]!=s[j] for i, j in zip(range(len(s)-1),range(1, len(s)))]):
        return s
    else:
        for i, j in zip(range(len(s)-1),range(1, len(s))):
            if s[i]==s[j]:
                s = s[:i] + s[j+1:]
                return superReducedString(s)

  • robertbielicki
     + 0 comments

    RUST: 

    fn superReducedString(s: &str) -> String {
    let mut last_idx = s.len() as i32;
    let mut result: Vec<&str> = s.split("").collect();
    let mut flag = true;

    while flag {
        flag = false;
        for n in 0..last_idx {
            if n <= last_idx {
                let first = n as usize;
                if result[first] == result[(n+1) as usize] {
                    result.remove(first);
                    result.remove(first);
                    last_idx -=2;
                    flag = true;
                } 
            }   
        }
    }
    if !(result.join("").is_empty()) {
        return result.join("")
    }
    "Empty String".to_string()
}