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> |
abababcdcdcdcd
Time Complexity: O(N*N)
Auxiliary Space: O(1)
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