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

Super Reduced String

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1710 Discussions

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  • zaidnsour12
     + 0 comments

    Solution using Stack in java:

     public static String superReducedString(String s) {
        String result = "";
        Stack<Character> stack = new Stack<>();
        int n = s.length();
        for(int i = 0; i < n; i++){
            char c = s.charAt(i);
            if(stack.empty() || c != stack.peek())
                stack.push(c);
            else
                stack.pop();
            }
        if(stack.empty())
            result = "Empty String";
        
        else
          for(char c: stack)
              result += c;
        
        return result;        
        }

  • 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;
}

  • amanrehanahmed
     + 0 comments
    #include <bits/stdc++.h>
#include <string>
using namespace std;

/*
 * Complete the 'superReducedString' function below.
 *
 * The function is expected to return a STRING.
 * The function accepts STRING s as parameter.
 */

string superReducedString(string s) {
    for(int i=1;i<s.length();i++){
        if(s[i]==s[i-1]){
            s.erase(s.begin()+i);
            s.erase((s.begin()+(i-1)));
        i=1;
            
        }
        if(s.length()==0){
            return "Empty String";
        }
        
    }
    if(s[0]==s[1]){
        return "Empty String";
    }
    return s;
}

int main()
{
    ofstream fout(getenv("OUTPUT_PATH"));

    string s;
    getline(cin, s);

    string result = superReducedString(s);

    fout << result << "\n";

    fout.close();

    return 0;
}

  • dhihanlaksanaaj1
     + 0 comments

    SWIFT

    func logic(charCount: Int, strings: [UInt8], string: UInt8) -> [UInt8] {
    var newStrings: [UInt8] = strings
    
    if charCount % 2 == 1 {
        if newStrings.last ?? 0 == string { newStrings.popLast() }
        else { newStrings.append(string) }
    }

    return newStrings
}

func superReducedString(s: String) -> String {
    // Write your code here
    var oldStrings: [UInt8] = [UInt8](s.utf8)
    var newStrings: [UInt8] = [UInt8]()
    
    var strTemp: UInt8 = oldStrings.first!
    var charCount: Int = 0
        
    for index in 0 ... oldStrings.count - 1 {
        if oldStrings[index] == strTemp {
          charCount += 1
          continue
        }
        
        newStrings = logic(charCount: charCount, strings: newStrings, string: oldStrings[index - charCount])
        strTemp = oldStrings[index]
        charCount = 1
    }
        
    newStrings = logic(charCount: charCount, strings: newStrings, string: oldStrings[oldStrings.count - 1])
    
    return newStrings.count == 0 ? "Empty String" : String(bytes: newStrings, encoding: String.Encoding.utf8) ?? ""
}

  • dotaikhoa
     + 0 comments

    c++ solution without erasing or adding stuff to data structures - using previous left and right (pl and pr) indexes and an array to store elements that have been "erased" (used vector).

    string superReducedString(string s) { int n = s.size(); vector used(n,false); int pl = -1, pr = -1;

    int i = 0;
while (i < n) {
    if (i+1 < n and s[i]==s[i+1]) {
        int l = !used[i] ? i : pl;
        int r = !used[i+1] ? i+1 : pr;

        while (l >= 0 and r < n) {
            if (s[l]!=s[r]) break;

            used[l]=used[r]=true;

            while (l>=0 and used[l]) l--;
            while (r<n and used[r]) r++;
        }
        pl = l, pr = r;
    }
    i=max(i+1,pr);
}