Decode a string recursively encoded as count followed by substring

Difficulty Level : Medium
Last Updated : 23 Apr, 2020

An encoded string (s) is given, the task is to decode it. The pattern in which the strings are encoded is as follows.

<count>[sub_str] ==> The substring 'sub_str' 
                      appears count times.

Examples:

Input : str[] = "1[b]"
Output : b

Input : str[] = "2[ab]"
Output : abab

Input : str[] = "2[a2[b]]"
Output : abbabb

Input : str[] = "3[b2[ca]]"
Output : bcacabcacabcaca

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is to use two stacks, one for integers and another for characters.
Now, traverse the string,

Whenever we encounter any number, push it into the integer stack and in case of any alphabet (a to z) or open bracket (‘[‘), push it onto the character stack.
Whenever any close bracket (‘]’) is encounter pop the character from the character stack until open bracket (‘[‘) is not found in the character stack. Also, pop the top element from the integer stack, say n. Now make a string repeating the popped character n number of time. Now, push all character of the string in the stack.

Below is implementation of this approach:

C++

// C++ program to decode a string recursively 
// encoded as count followed substring 
#include<bits/stdc++.h> 
using namespace std; 
  
// Returns decoded string for 'str' 
string decode(string str) 
{ 
    stack<int> integerstack; 
    stack<char> stringstack; 
    string temp = "", result = ""; 
  
    // Traversing the string 
    for (int i = 0; i < str.length(); i++) 
    { 
        int count = 0; 
  
        // If number, convert it into number 
        // and push it into integerstack. 
        if (str[i] >= '0' && str[i] <='9') 
        { 
            while (str[i] >= '0' && str[i] <= '9') 
            { 
                count = count * 10 + str[i] - '0'; 
                i++; 
            } 
  
            i--; 
            integerstack.push(count); 
        } 
  
        // If closing bracket ']', pop elemment until 
        // '[' opening bracket is not found in the 
        // character stack. 
        else if (str[i] == ']') 
        { 
            temp = ""; 
            count = 0; 
  
            if (! integerstack.empty()) 
            { 
                count = integerstack.top(); 
                integerstack.pop(); 
            } 
  
            while (! stringstack.empty() && stringstack.top()!='[' ) 
            { 
                temp = stringstack.top() + temp; 
                stringstack.pop(); 
            } 
  
            if (! stringstack.empty() && stringstack.top() == '[') 
                stringstack.pop(); 
  
            // Repeating the popped string 'temo' count 
            // number of times. 
            for (int j = 0; j < count; j++) 
                result = result + temp; 
  
            // Push it in the character stack. 
            for (int j = 0; j < result.length(); j++) 
                stringstack.push(result[j]); 
  
            result = ""; 
        } 
  
        // If '[' opening bracket, push it into character stack. 
        else if (str[i] == '[') 
        { 
            if (str[i-1] >= '0' && str[i-1] <= '9') 
                stringstack.push(str[i]); 
  
            else
            { 
                stringstack.push(str[i]); 
                integerstack.push(1); 
            } 
        } 
  
        else
            stringstack.push(str[i]); 
    } 
  
    // Pop all the elmenet, make a string and return. 
    while (! stringstack.empty()) 
    { 
        result = stringstack.top() + result; 
        stringstack.pop(); 
    } 
  
    return result; 
} 
  
// Driven Program 
int main() 
{ 
    string str = "3[b2[ca]]"; 
    cout << decode(str) << endl; 
    return 0; 
} 

Java

// Java program to decode a string recursively 
// encoded as count followed substring 
  
import java.util.Stack; 
  
class Test 
{ 
    // Returns decoded string for 'str' 
    static String decode(String str) 
    { 
        Stack<Integer> integerstack = new Stack<>(); 
        Stack<Character> stringstack = new Stack<>(); 
        String temp = "", result = ""; 
       
        // Traversing the string 
        for (int i = 0; i < str.length(); i++) 
        { 
            int count = 0; 
       
            // If number, convert it into number 
            // and push it into integerstack. 
            if (Character.isDigit(str.charAt(i))) 
            { 
                while (Character.isDigit(str.charAt(i))) 
                { 
                    count = count * 10 + str.charAt(i) - '0'; 
                    i++; 
                } 
       
                i--; 
                integerstack.push(count); 
            } 
       
            // If closing bracket ']', pop elemment until 
            // '[' opening bracket is not found in the 
            // character stack. 
            else if (str.charAt(i) == ']') 
            { 
                temp = ""; 
                count = 0; 
       
                if (!integerstack.isEmpty()) 
                { 
                    count = integerstack.peek(); 
                    integerstack.pop(); 
                } 
       
                while (!stringstack.isEmpty() && stringstack.peek()!='[' ) 
                { 
                    temp = stringstack.peek() + temp; 
                    stringstack.pop(); 
                } 
       
                if (!stringstack.empty() && stringstack.peek() == '[') 
                    stringstack.pop(); 
       
                // Repeating the popped string 'temo' count 
                // number of times. 
                for (int j = 0; j < count; j++) 
                    result = result + temp; 
       
                // Push it in the character stack. 
                for (int j = 0; j < result.length(); j++) 
                    stringstack.push(result.charAt(j)); 
       
                result = ""; 
            } 
       
            // If '[' opening bracket, push it into character stack. 
            else if (str.charAt(i) == '[') 
            { 
                if (Character.isDigit(str.charAt(i-1))) 
                    stringstack.push(str.charAt(i)); 
       
                else
                { 
                    stringstack.push(str.charAt(i)); 
                    integerstack.push(1); 
                } 
            } 
       
            else
                stringstack.push(str.charAt(i)); 
        } 
       
        // Pop all the elmenet, make a string and return. 
        while (!stringstack.isEmpty()) 
        { 
            result = stringstack.peek() + result; 
            stringstack.pop(); 
        } 
       
        return result; 
    } 
  
    // Driver method 
    public static void main(String args[]) 
    { 
        String str = "3[b2[ca]]"; 
        System.out.println(decode(str)); 
    } 
} 

Python3

# Python program to decode a string recursively  
# encoded as count followed substring  
  
# Returns decoded string for 'str'  
def decode(Str): 
    integerstack = [] 
    stringstack = [] 
    temp = "" 
    result = ""  
    i = 0
    # Traversing the string  
    while i < len(Str): 
        count = 0
  
        # If number, convert it into number  
        # and push it into integerstack.  
        if (Str[i] >= '0' and Str[i] <='9'): 
            while (Str[i] >= '0' and Str[i] <= '9'): 
                count = count * 10 + ord(Str[i]) - ord('0')  
                i += 1
            i -= 1
            integerstack.append(count) 
  
        # If closing bracket ']', pop elemment until  
        # '[' opening bracket is not found in the  
        # character stack.  
        elif (Str[i] == ']'): 
            temp = ""  
            count = 0
  
            if (len(integerstack) != 0): 
                count = integerstack[-1]  
                integerstack.pop() 
  
            while (len(stringstack) != 0 and stringstack[-1] !='[' ): 
                temp = stringstack[-1] + temp  
                stringstack.pop() 
  
            if (len(stringstack) != 0 and stringstack[-1] == '['):  
                stringstack.pop()  
  
            # Repeating the popped string 'temo' count  
            # number of times. 
            for j in range(count): 
                result = result + temp  
  
            # Push it in the character stack. 
            for j in range(len(result)): 
                stringstack.append(result[j])  
  
            result = "" 
  
        # If '[' opening bracket, push it into character stack.  
        elif (Str[i] == '['): 
            if (Str[i-1] >= '0' and Str[i-1] <= '9'):  
                stringstack.append(Str[i])  
  
            else: 
                stringstack.append(Str[i])  
                integerstack.append(1) 
  
        else: 
            stringstack.append(Str[i]) 
          
        i += 1
  
    # Pop all the elmenet, make a string and return. 
    while len(stringstack) != 0: 
        result = stringstack[-1] + result  
        stringstack.pop() 
  
    return result 
  
# Driven code  
if __name__ == '__main__': 
    Str = "3[b2[ca]]"
    print(decode(Str)) 
      
# This code is contributed by PranchalK. 

C#

// C# program to decode a string recursively  
// encoded as count followed substring  
using System; 
using System.Collections.Generic; 
  
class GFG 
{ 
// Returns decoded string for 'str'  
public static string decode(string str) 
{ 
    Stack<int> integerstack = new Stack<int>(); 
    Stack<char> stringstack = new Stack<char>(); 
    string temp = "", result = ""; 
  
    // Traversing the string  
    for (int i = 0; i < str.Length; i++) 
    { 
        int count = 0; 
  
        // If number, convert it into number  
        // and push it into integerstack.  
        if (char.IsDigit(str[i])) 
        { 
            while (char.IsDigit(str[i])) 
            { 
                count = count * 10 + str[i] - '0'; 
                i++; 
            } 
  
            i--; 
            integerstack.Push(count); 
        } 
  
        // If closing bracket ']', pop elemment  
        // until '[' opening bracket is not found  
        // in the character stack.  
        else if (str[i] == ']') 
        { 
            temp = ""; 
            count = 0; 
  
            if (integerstack.Count > 0) 
            { 
                count = integerstack.Peek(); 
                integerstack.Pop(); 
            } 
  
            while (stringstack.Count > 0 &&  
                   stringstack.Peek() != '[') 
            { 
                temp = stringstack.Peek() + temp; 
                stringstack.Pop(); 
            } 
  
            if (stringstack.Count > 0 &&  
                stringstack.Peek() == '[') 
            { 
                stringstack.Pop(); 
            } 
  
            // Repeating the popped string 'temo'  
            // count number of times.  
            for (int j = 0; j < count; j++) 
            { 
                result = result + temp; 
            } 
  
            // Push it in the character stack.  
            for (int j = 0; j < result.Length; j++) 
            { 
                stringstack.Push(result[j]); 
            } 
  
            result = ""; 
        } 
  
        // If '[' opening bracket, push it  
        // into character stack.  
        else if (str[i] == '[') 
        { 
            if (char.IsDigit(str[i - 1])) 
            { 
                stringstack.Push(str[i]); 
            } 
  
            else
            { 
                stringstack.Push(str[i]); 
                integerstack.Push(1); 
            } 
        } 
  
        else
        { 
            stringstack.Push(str[i]); 
        } 
    } 
  
    // Pop all the elmenet, make a  
    // string and return.  
    while (stringstack.Count > 0) 
    { 
        result = stringstack.Peek() + result; 
        stringstack.Pop(); 
    } 
  
    return result; 
} 
  
// Driver Code  
public static void Main(string[] args) 
{ 
    string str = "3[b2[ca]]"; 
    Console.WriteLine(decode(str)); 
} 
} 
  
// This code is contributed by Shrikant13 

Output:

bcacabcacabcaca

<!—-Illustration of above code for “3[b2[ca]]”>
<!—->

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Related Articles

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#