# HackerRank Super Reduced String problem solution

In this HackerRank Super Reduced String problem, you need to 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

## Problem solution in Python programming.

```s = input()

changed = True
while changed and s != "":
changed = False
for i in range(len(s) - 1):
if s[i] == s[i+1]:
changed = True
s = s[:(i)] + s[(i+2):]
break

if s == "":
print('Empty String')
else:
print(s)```

## Problem solution in Java Programming.

```import java.io.*;
import java.util.*;

public class Solution {

public static void main(String [] args){
try {
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Scanner in = new Scanner(fr);*/
Scanner in = new Scanner(System.in);
reduceString(in.nextLine());

}

static void reduceString(String s) {
boolean found = false;
for (int i = 0; i < s.length() - 1; i++) {
String a = s.substring(i, i + 1);
String b = s.substring(i + 1, i + 2);
if (a.equals(b)) {
s = s.substring(0, i) + s.substring(i + 2, s.length());

reduceString(s);
found = true;
break;
}
}
if(!found){
if(s.isEmpty()){
System.out.println("Empty String");
} else {
System.out.println(s);
}
}

}

}```

### Problem solution in C++ programming.

```#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;

char s[107];

bool redukuj() {
int n = 0;
for(; s[n]; ++n);
for(int i = 0; s[i + 1]; ++i) {
if(s[i] == s[i + 1]) {
for(int j = i; s[j + 2]; ++j) {
s[j] = s[j + 2];
}
s[n - 1] = s[n - 2] = 0;
return true;
}
}
return false;
}

int main() {
ios_base::sync_with_stdio(0);
cin >> s;
while(redukuj());
if(!s[0]) cout << "Empty String\n";
else cout << s << endl;
return 0;
}```

### Problem solution in C programming.

```#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define MAX_LENGTH 101

int main() {

/* Enter your code here. Read input from STDIN. Print output to STDOUT */
char str[MAX_LENGTH];
fgets(str,MAX_LENGTH,stdin);

int length = strlen(str);

if(length < 2){
printf("Empty String\n");
return 0;
}

bool done = false;
while(!done){
done = true;
if(strlen(str) < 2) break;

for(int i =0; i < strlen(str)-1; i++){
if(str[i] ==  str[i+1]){
done = false;
memmove(&str[i],&str[i+2],strlen(str)-i-1);
break;
}
}
}
if(strlen(str)){
printf("%s\n",str);
return 0;
}
printf("Empty String\n");
return 0;
}```

### Problem solution in JavaScript programming.

```function processData(input) {
let arr = input.split('');
for(let i = 0; i< arr.length; ++i) {
if(arr[i] === arr[i+1]) {
arr.splice(i, 2);
i = -1;
}
}
if(arr.length === 0) console.log('Empty String');
console.log(arr.join(''));
}

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

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

1. the pytho solution doesn' work for me

1. changed = True
while changed and s != "":
changed = False
for i in range(1,len(s) - 1):
if s[i] == s[i+1]:
changed = True
s = s[:(i)] + s[(i+2):]
break
print(s)
if s[0] == s[1]:
s = s[2:]
return s if s else 'Empty String'