HackerRank Super Reduced String Solution

In this post, we will solve HackerRank Super Reduced String Problem Solution.

Reduce a string of lowercase characters in range ascii [‘a’..’z’] by doing a series of operations. In each operation, select a pair of adjacent letters that match, and delete them. Delete as many characters as possible using this method and return the resulting string. If the final string is empty, return Empty String
Example
s = ‘aab’
aab shortens to b in one operation: remove the adjacent a characters.
s = ‘abba’
Remove the two ‘b’ characters leaving ‘aa’. Remove the two ‘a’ characters to leave “. Return ‘Empty String’.

Function Description

Complete the superReducedString function in the editor below.

superReducedString has the following parameter(s):

string s: a string to reduce

Returns

string: the reduced string or Empty String

Input Format

A single string, s

Sample Input 0

aaabccddd

Sample Output 0

abd

Explanation 0

Perform the following sequence of operations to get the final string:

aaabccddd → abccddd → abddd → abd

Sample Input 1

aa

Sample Output 1

Empty String

Explanation 1

aa → Empty String

Sample Input 2

baab

Sample Output 2

Empty String

Explanation 2

baab → bb → Empty String

HackerRank Super Reduced String Problem Solution

Super Reduced String C Solution

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    char s[103];
    
    scanf("%s", s);
    
    int len, ll;
    do {
        len = strlen(s);
//        printf("%d\n", len);
        for (char *ss=s; *ss; ss++) {
            if (*ss == ss[1]) {
//                printf("%s ", s );
                strcpy(ss, ss+2);
//                printf("%s\n", s );
            } else {
//                printf("\n");
            }   
        }
        ll = strlen(s);
//        printf("%d\n", ll);
    } while (len != ll);
    
    
    if (s[0]==0) {
        printf("Empty String\n");
    } else {
        printf("%s\n", s);
    }
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}

Super Reduced String C++ Solution

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define all(c) (c).begin(), (c).end()
#define mp make_pair

string s, tmp = "";
int c = 0;	
char str;
bool vis[105];

int main(){
	ios_base :: sync_with_stdio(0);
	cin >> s;
	while(1){
		if(c++ == 10000) break;
		for(int i = 1; i < (int)s.size(); i++){
			if((!vis[i]) && (!vis[i-1])){
				if(s[i] == s[i-1]) vis[i] = 1, vis[i-1] = 1;
			}
		}	
		for(int i = 0; i < (int)s.size(); i++){
			if(!vis[i]) tmp += s[i];
		}
		s = tmp; tmp = ""; memset(vis, 0, sizeof vis);
	}
	if(s == "") cout << "Empty String" << '\n';
	else cout << s << '\n';	
	return 0;
}

Super Reduced String C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
class Solution {
    static void Main(String[] args) 
    {
      string inputStr = Console.ReadLine();
      
      
      while(true)
      {
        bool found = false;
        
        for(int i = 0 ; i < inputStr.Length - 1; i++)
        {
          if( inputStr[i]==inputStr[i+1] )
          {
            inputStr = inputStr.Remove(i,2);
            found = true;
            continue;
          }
        }
        
        if(!found)
          break;
      }
      if(inputStr.Length == 0)
        Console.WriteLine("Empty String");  
      else
      Console.WriteLine(inputStr);
    }
}

Super Reduced String Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

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

    public static String superReducedString(String s) {
      if(s.length()==1){
            return s;
        }
        char[] chars = s.toCharArray();
        LinkedList<Character> charStack =  new LinkedList<>();
        for (int i=0;i<chars.length;i++){
            Character last = charStack.peekLast();
            if(last != null && last.equals(chars[i])){
                charStack.removeLast();
            }else{
                charStack.addLast(chars[i]);
            }
        }
        // print charStackContent
        if(charStack.isEmpty()){
            return "Empty String";
        }
        return charStack.stream().map(c->c.toString()).collect(Collectors.joining());
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        String s = bufferedReader.readLine();

        String result = Result.superReducedString(s);

        bufferedWriter.write(result);
        bufferedWriter.newLine();

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Super Reduced String JavaScript Solution

function processData(input) {
    var s = input.concat(); 
    
    for(i=0; i < s.length-1; )
    {
        if(s[i] === s[i+1])
        {
            s = s.slice(0, i) + s.slice(i+2); 
            i = 0; 
        }
        else
        {
            i++; 
        }
    }

    console.log(s || "Empty String"); 
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});

Super Reduced String Python Solution

s = input()
st = []
c = 0
cur = s[0]
for i in s:
    if i == cur:
         c += 1
    else:
        if c%2 != 0:
            if st:
                if st[-1] == cur:
                    st.pop(-1)
                else:
                    st.append(cur)
            else:
                st.append(cur)
        cur = i
        c = 1
if c%2 != 0:
    if st:
        if st[-1] == cur:
            st.pop(-1)
        else:
            st.append(cur)
    else:
         st.append(cur)
s = ''.join(st)
if not s:
    s = "Empty String"
print(s)