Hello Programmers, In this post, you will know how to solve the HackerRank Super Reduced String Solution. This problem is a part of the HackerRank Algorithms Series.
One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.
HackerRank Super Reduced String Solution
Task
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.
Constraints
- 1 <= length of s <= 100
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 Solutions
Super Reduced String Solution in C
#include<stdio.h>
typedef unsigned u;
char S[111];u l=1;
int main()
{
char c;
while((c=getchar())<'a');
do
{
S[l]=c;
if(S[l]==S[l-1])--l;
else++l;
}
while((c=getchar())>='a');
S[l]='\0';
printf("%s\n",l-1?S+1:"Empty String");
return 0;
}Super Reduced String Solution in Cpp
#include <iostream>
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<cstdio>
#include<numeric>
#include<cstring>
#include<ctime>
#include<cstdlib>
#include<set>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<list>
#include<cmath>
#include<bitset>
#include<cassert>
#include<queue>
#include<stack>
#include<deque>
#include<cassert>
using namespace std;
typedef long long ll;
typedef long double ld;
int main()
{
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
string s;
cin >> s;
vector<char>st;
for (int i = 0; i < (int)s.length(); i++)
{
if (!st.empty() && st.back() == s[i])
{
st.pop_back();
}
else
{
st.push_back(s[i]);
}
}
if (st.empty())
{
printf("Empty String\n");
}
else
{
for (int i = 0; i < (int)st.size(); i++)
{
printf("%c", st[i]);
}
printf("\n");
}
return 0;
}
Super Reduced String Solution in Java
import java.awt.*;
import java.awt.event.*;
import java.awt.geom.*;
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;
/*
br = new BufferedReader(new FileReader("input.txt"));
pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
br = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
*/
public class Solution {
private static BufferedReader br;
private static StringTokenizer st;
private static PrintWriter pw;
public static void main(String[] args) throws Exception {
br = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
//int qq = 1;
int qq = Integer.MAX_VALUE;
//int qq = readInt();
for(int casenum = 1; casenum <= qq; casenum++) {
String s = nextToken();
LinkedList<Character> q = new LinkedList<Character>();
for(int i = 0; i < s.length(); i++) {
if(!q.isEmpty() && q.peekLast() == s.charAt(i)) {
q.removeLast();
}
else {
q.addLast(s.charAt(i));
}
}
for(char out: q) {
pw.print(out);
}
if(q.size() == 0) {
pw.print("Empty String");
}
pw.println();
}
exitImmediately();
}
private static void exitImmediately() {
pw.close();
System.exit(0);
}
private static long readLong() throws IOException {
return Long.parseLong(nextToken());
}
private static double readDouble() throws IOException {
return Double.parseDouble(nextToken());
}
private static int readInt() throws IOException {
return Integer.parseInt(nextToken());
}
private static String nextLine() throws IOException {
if(!br.ready()) {
exitImmediately();
}
st = null;
return br.readLine();
}
private static String nextToken() throws IOException {
while(st == null || !st.hasMoreTokens()) {
if(!br.ready()) {
exitImmediately();
}
st = new StringTokenizer(br.readLine().trim());
}
return st.nextToken();
}
}Super Reduced String Solution in Python
# Enter your code here. Read input from STDIN. Print output to STDOUT
s = raw_input()
while 1:
i=0
flag=0
while i<len(s)-1:
if s[i] == s[i+1]:
s=s[:i]+s[i+2:]
flag=1
else:
i+=1
if flag==0:
break
if len(s)>0:
print s
else:
print 'Empty String'Super Reduced String Solution using JavaScript
function processData(input) {
//Enter your code here
var continueReducing = true;
var s = input.split('');
while(continueReducing) {
continueReducing = false;
for(var i = 0; i < s.length; i++) {
if (s[i] == s[i + 1]) {
s[i] = '';
s[i + 1] = '';
i++;
continueReducing = true;
}
}
s = s.join('').split('');
}
console.log(s.length == 0 ? 'Empty String' : s.join(''));
}
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 Solution in Scala
object Solution {
val pairs = (('a' to 'z') zip ('a' to 'z')).map(s => "" + s._1 + s._2).toList
def main(args: Array[String]) {
val sc = new java.util.Scanner(System.in)
//println(pairs.mkString(" "))
val res = reduce(sc.next)
if (res.length == 0)
println("Empty String")
else
println(res)
}
def reduce(str: String) = {
def step(s: String) = {
var aux = s
for (pair <- pairs) {
aux = aux.replaceAll(pair, "")
}
aux
}
var pre = str
var post = step(pre)
while (pre.length != post.length) {
pre = post
post = step(pre)
}
post
}
}Super Reduced String Solution in Pascal
program test;
uses sysutils;
var s,al,s1: string;
i,j: integer;
NotFound: boolean;
begin
al := '';
readln(s);
s:=ansilowercase(s);
NotFound := False;
for i:=1 to length(s) do
begin
s1 := s[i];
//writeln(al + ' ' + s1);
if (Length(al)>0) then
begin
if al[Length(al)] = s1 then
begin
Delete(al, Length(al), 1);
end
else
begin
al := al + s1;
end;
end
else
begin
al := al + s1;
end;
end;
if Length(al)=0 then
writeln('Empty String')
else
writeln(al);
end.Disclaimer: This problem (Super Reduced String) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.
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