Expand the string according to the given conditions

Given a string str of the type “3(ab)4(cd)”, the task is to expand it to “abababcdcdcdcd” where integers are from the range [1, 9].

This problem was asked in ThoughtWorks interview held in October 2018.

Examples:



Input: str = “3(ab)4(cd)”
Output: abababcdcdcdcd

Input: str = “2(kl)3(ap)”
Output: klklapapap

Approach: We traverse through the string and wait for a numeric value, num to turn up at position i. As soon as it arrives, we check i + 1 for a ‘(‘. If it’s present then program enters into a loop to extract whatever is within ‘(‘ and ‘)’ and concatenate it to an empty string, temp. Later another loop prints the generated string num number of times. Repeat these steps until the string finishes.

Below is the implementation of the approach:

C++

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// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
  
// Function to expand and print the given string
void expandString(string strin)
{
    string temp = "";
    int j;
  
    for (int i = 0; i < strin.length(); i++) 
    {
        if (strin[i] >= 0) 
        {
  
            // Subtract '0' to convert char to int
            int num = strin[i] - '0';
            if (strin[i + 1] == '(') 
            {
  
                // Characters within brackets
                for (j = i + 1; strin[j] != ')'; j++)
                {
                    if ((strin[j] >= 'a' && strin[j] <= 'z') || 
                        (strin[j] >= 'A' && strin[j] <= 'Z'))
                    {
                        temp += strin[j];
                    }
                }
  
                // Expanding
                for (int k = 1; k <= num; k++) 
                {
                    cout << (temp);
                }
  
                // Reset the variables
                num = 0;
                temp = "";
  
                if (j < strin.length()) 
                {
                    i = j;
                }
            }
        }
    }
}
  
// Driver code
int main()
{
    string strin = "3(ab)4(cd)";
    expandString(strin);
}
  
// This code is contributed by Surendra_Gangwar

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// Java implementation of the approach 


public class GFG { 


  


    // Function to expand and print the given string 


    static void expandString(String strin) 


    { 


        String temp = ""; 


        int j; 


  


        for (int i = 0; i < strin.length(); i++) { 


            if (strin.charAt(i) >= 0) { 


  


                // Subtract '0' to convert char to int 


                int num = strin.charAt(i) - '0'; 


                if (strin.charAt(i + 1) == '(') { 


  


                    // Characters within brackets 


                    for (j = i + 1; strin.charAt(j) != ')'; j++) { 


                        if ((strin.charAt(j) >= 'a'


                             && strin.charAt(j) <= 'z') 


                            || (strin.charAt(j) >= 'A'


                                && strin.charAt(j) <= 'Z')) { 


                            temp += strin.charAt(j); 


                        } 


                    } 


  


                    // Expanding 


                    for (int k = 1; k <= num; k++) { 


                        System.out.print(temp); 


                    } 


  


                    // Reset the variables 


                    num = 0; 


                    temp = ""; 


  


                    if (j < strin.length()) { 


                        i = j; 


                    } 


                } 


            } 


        } 


    } 


  


    // Driver code 


    public static void main(String args[]) 


    { 


        String strin = "3(ab)4(cd)"; 


        expandString(strin); 


    } 


} 



















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Output:


abababcdcdcdcd






          
          
          
          

            

    

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Improved By :  SURENDRA_GANGWAR