HackerRank Super Reduced String problem solution

HackerRank Super Reduced String problem solution – In this HackerRank Super Reduced String problem, 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.

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.

HackerRank Super Reduced String problem solution in Python.

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)

Super Reduced String problem solution in Java.

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

public class Solution {
	
	public static void main(String [] args){
		/*FileReader fr = null;
		try {
			fr = new FileReader("/Users/pkahrl/Documents/eclipse-workspace/Comp/src/kahrl/prob1/input.txt");
		} 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 Stringn";
	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 Stringn");
        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("%sn",str);
      return 0;
    } 
    printf("Empty Stringn");
    return 0;
}

Problem solution in JavaScript programming.

function processData(input) {
    //Enter your code here
  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);
});