Super Reduced String


Problem Statement :


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.

Constraints

1  <=  len of s  <=  100


Sample Input 0

aaabccddd
Sample Output 0

abd


Solution :



title-img 


                            Solution in C :

In   C++   :







#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;
}








In    Java   :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
      
        Scanner in = new Scanner(System.in);
        String S = in.nextLine();
        boolean changed = false;
        do{
            changed = false;
            for (int i = 0; i < S.length(); i++){
                if (i == S.length() - 1) continue;
                if (S.charAt(i) == S.charAt(i+1)){
                    changed = true;
                    S = S.substring(0,i) + S.substring(i+2);
                }
            }
        }while(changed);
        if (S.equals("")) System.out.println("Empty String");
        else System.out.println(S);
    }
}








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;
}








In   Python3 :






s = input()
while True:
    for i in range(len(s)-1):
        if s[i] == s[i+1]:
            s = s[:i]+s[i+2:]
            break
    else:
        break
print(s if s else "Empty String")








View More Similar Problems

Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

View Solution →

Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

View Solution →

Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty

View Solution →

Median Updates

The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o

View Solution →

Maximum Element

You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each

View Solution →

Balanced Brackets

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra

View Solution →