Expand the string according to the given conditions

Given 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 the 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++

// 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

Java

// 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);
    }
}

Python3

# Python3 implementation of the approach

# Function to expand and print the given string
def expandString(strin):

    temp = ""
    j = 0
    i = 0
    while(i < len(strin)):
        if (strin[i] >= "0"):

            # Subtract '0' to convert char to int
            num = ord(strin[i])-ord("0")
            if (strin[i + 1] == '('):

                # Characters within brackets
                j = i + 1
                while(strin[j] != ')'):
                    if ((strin[j] >= 'a' and strin[j] <= 'z') or \
                        (strin[j] >= 'A' and strin[j] <= 'Z')):
                        temp += strin[j]
                    j += 1

                # Expanding
                for k in range(1, num + 1):
                    print(temp,end="")

                # Reset the variables
                num = 0
                temp = ""
                if (j < len(strin)):
                    i = j
        i += 1

# Driver code
strin = "3(ab)4(cd)"
expandString(strin)

# This code is contributed by shubhamsingh10

C#

// C# implementation of
// the above approach
using System;
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[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++)
        {
          Console.Write(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);
}
}

// This code is contributed by Chitranayal

Javascript

<script>
// Javascript implementation of the approach

    // Function to expand and print the given string
    function expandString(strin)
    {
        let temp = "";
        let j;

        for (let i = 0; i < strin.length; i++) {
            if (strin[i].charCodeAt(0) >= 0) {

                // Subtract '0' to convert char to int
                let num = strin[i].charCodeAt(0) - '0'.charCodeAt(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 (let k = 1; k <= num; k++) {
                        document.write(temp);
                    }

                    // Reset the variables
                    num = 0;
                    temp = "";

                    if (j < strin.length) {
                        i = j;
                    }
                }
            }
        }
    }

    // Driver code
    let strin = "3(ab)4(cd)";
    expandString(strin);


// This code is contributed by rag2127
</script>

Output:

abababcdcdcdcd

Time Complexity: O(N*N)
Auxiliary Space: O(1)