Decode a string recursively encoded as count followed by substring

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

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

1. 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.
2. 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 the 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 element 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 element, 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 element 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 element, 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 element 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 element, 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 element
        // 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 element, 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

Javascript

<script>
    // Javascript program to decode a string recursively
    // encoded as count followed substring

    // Returns decoded string for 'str'
    function decode(str)
    {
        let integerstack = [];
        let stringstack = [];
        let temp = "", result = "";

        // Traversing the string
        for (let i = 0; i < str.length; i++)
        {
            let 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 element
            // until '[' opening bracket is not found
            // in the character stack.
            else if (str[i] == ']')
            {
                temp = "";
                count = 0;

                if (integerstack.length > 0)
                {
                    count = integerstack[integerstack.length - 1];
                    integerstack.pop();
                }

                while (stringstack.length > 0 &&
                       stringstack[stringstack.length - 1] != '[')
                {
                    temp = stringstack[stringstack.length - 1] + temp;
                    stringstack.pop();
                }

                if (stringstack.length > 0 &&
                    stringstack[stringstack.length - 1] == '[')
                {
                    stringstack.pop();
                }

                // Repeating the popped string 'temo'
                // count number of times.
                for (let j = 0; j < count; j++)
                {
                    result = result + temp;
                }

                // Push it in the character stack.
                for (let 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 element, make a
        // string and return.
        while (stringstack.length > 0)
        {
            result = stringstack[stringstack.length - 1] + result;
            stringstack.pop();
        }

        return result;
    }

    let str = "3[b2[ca]]";
    document.write(decode(str));

// This code is contributed by divyeshrabadiy07.
</script>

Output

bcacabcacabcaca

Time Complexity: O(n)
Auxiliary Space: O(n)

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

Method 2(Using 1 stack)

Algorithm:
Loop through the characters of the string
If the character is not ']', add it to the stack
If the character is ']':
   While top of the stack doesn't contain '[', pop the characters from the stack and store it in a string temp
   (Make sure the string isn't in reverse order)
   Pop '[' from the stack
   While the top of the stack contains a digit, pop it and store it in dig
   Concatenate the string temp for dig number of times and store it in a string repeat
   Add the string repeat to the stack
Pop all the characters from the stack(also make the string isn't in reverse order) 

Below is the implementation of this approach: 

C++

#include <iostream>
#include <stack>
using namespace std;

string decodeString(string s)
{
    stack<char> st;
    for (int i = 0; i < s.length(); i++) {
        // When ']' is encountered, we need to start
        // decoding
        if (s[i] == ']') {
            string temp;
            while (!st.empty() && st.top() != '[') {
                // st.top() + temp makes sure that the
                // string won't be in reverse order eg, if
                // the stack contains 12[abc temp = c + "" =>
                // temp = b + "c" => temp = a + "bc"
                temp = st.top() + temp;
                st.pop();
            }
            // remove the '[' from the stack
            st.pop();
            string num;
            // remove the digits from the stack
            while (!st.empty() && isdigit(st.top())) {
                num = st.top() + num;
                st.pop();
            }
            int number = stoi(num);
            string repeat;
            for (int j = 0; j < number; j++)
                repeat += temp;
            for (char c : repeat)
                st.push(c);
        }
        // if s[i] is not ']', simply push s[i] to the stack
        else
            st.push(s[i]);
    }
    string res;
    while (!st.empty()) {
        res = st.top() + res;
        st.pop();
    }
    return res;
}
// driver code
int main()
{
    string str = "3[b2[ca]]";
    cout << decodeString(str);
    return 0;
}

Java

import java.util.*;
public class Main
{
    static String decodeString(String s)
    {
        Vector<Character> st = new Vector<Character>();

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

            // When ']' is encountered, we need to
            // start decoding
            if (s.charAt(i) == ']')
            {
                String temp = "";
                while (st.size() > 0 && st.get(st.size() - 1) != '[')
                {

                    // st.top() + temp makes sure that the
                    // string won't be in reverse order eg, if
                    // the stack contains 12[abc temp = c + "" =>
                    // temp = b + "c" => temp = a + "bc"
                    temp = st.get(st.size() - 1) + temp;
                    st.remove(st.size() - 1);
                }

                // Remove the '[' from the stack
                st.remove(st.size() - 1);
                String num = "";

                // Remove the digits from the stack
                while (st.size() > 0 &&
                       st.get(st.size() - 1) >= 48 &&
                       st.get(st.size() - 1) <= 57)
                {
                    num = st.get(st.size() - 1) + num;
                    st.remove(st.size() - 1);
                }

                int number = Integer.parseInt(num);
                String repeat = "";
                for(int j = 0; j < number; j++)
                    repeat += temp;

                for(int c = 0; c < repeat.length(); c++)
                    st.add(repeat.charAt(c));
            }

            // If s[i] is not ']', simply push
            // s[i] to the stack
            else
                st.add(s.charAt(i));
        }
        String res = "";
        while (st.size() > 0)
        {
            res = st.get(st.size() - 1) + res;
            st.remove(st.size() - 1);
        }
        return res;
    }

    public static void main(String[] args) {
        String str = "3[b2[ca]]";
        System.out.print(decodeString(str));
    }
}

// This code is contributed by suresh07.

Python3

def decodeString(s):
    st = []
    for i in range(len(s)):

        # When ']' is encountered, we need to start decoding
        if s[i] == ']':
            temp = ""
            while len(st) > 0 and st[-1] != '[':

                # st.top() + temp makes sure that the
                # string won't be in reverse order eg, if
                # the stack contains 12[abc temp = c + "" =>
                # temp = b + "c" => temp = a + "bc"
                temp = st[-1] + temp
                st.pop()

            # remove the '[' from the stack
            st.pop()
            num = ""

            # remove the digits from the stack
            while len(st) > 0 and ord(st[-1]) >= 48 and ord(st[-1]) <= 57:
                num = st[-1] + num
                st.pop()
            number = int(num)
            repeat = ""
            for j in range(number):
                repeat += temp
            for c in range(len(repeat)):
                if len(st) > 0:
                    if repeat == st[-1]:
                        break #otherwise this thingy starts appending the same decoded words
                st.append(repeat)

        else:
            st.append(s[i])

    return st[0]


Str = "3[b2[ca]]"
print(decodeString(Str))

# This code is contributed by mukesh07.
# And debugged by ivannakreshchenetska

C#

using System;
using System.Collections.Generic;

class GFG{

static string decodeString(string s)
{
    List<char> st = new List<char>();

    for(int i = 0; i < s.Length; i++)
    {

        // When ']' is encountered, we need to
        // start decoding
        if (s[i] == ']')
        {
            string temp = "";
            while (st.Count > 0 && st[st.Count - 1] != '[')
            {

                // st.top() + temp makes sure that the
                // string won't be in reverse order eg, if
                // the stack contains 12[abc temp = c + "" =>
                // temp = b + "c" => temp = a + "bc"
                temp = st[st.Count - 1] + temp;
                st.RemoveAt(st.Count - 1);
            }

            // Remove the '[' from the stack
            st.RemoveAt(st.Count - 1);
            string num = "";

            // Remove the digits from the stack
            while (st.Count > 0 &&
                   st[st.Count - 1] >= 48 &&
                   st[st.Count - 1] <= 57)
            {
                num = st[st.Count - 1] + num;
                st.RemoveAt(st.Count - 1);
            }

            int number = int.Parse(num);
            string repeat = "";
            for(int j = 0; j < number; j++)
                repeat += temp;

            foreach(char c in repeat)
                st.Add(c);
        }

        // If s[i] is not ']', simply push
        // s[i] to the stack
        else
            st.Add(s[i]);
    }
    string res = "";
    while (st.Count > 0)
    {
        res = st[st.Count - 1] + res;
        st.RemoveAt(st.Count - 1);
    }
    return res;
}

// Driver code
static void Main()
{
    string str = "3[b2[ca]]";
    Console.Write(decodeString(str));
}
}

// This code is contributed by decode2207

Javascript

<script>

    function decodeString(s)
    {
        let st = [];
        for (let i = 0; i < s.length; i++)
        {

            // When ']' is encountered, we need to start
            // decoding
            if (s[i] == ']') {
                let temp = "";
                while (st.length > 0 && st[st.length - 1] != '[')
                {

                    // st.top() + temp makes sure that the
                    // string won't be in reverse order eg, if
                    // the stack contains 12[abc temp = c + "" =>
                    // temp = b + "c" => temp = a + "bc"
                    temp = st[st.length - 1] + temp;
                    st.pop();
                }

                // remove the '[' from the stack
                st.pop();
                let num = "";

                // remove the digits from the stack
                while (st.length > 0 && st[st.length - 1].charCodeAt() >= 48 && st[st.length - 1].charCodeAt() <= 57) {
                    num = st[st.length - 1] + num;
                    st.pop();
                }
                let number = parseInt(num);
                let repeat = "";
                for (let j = 0; j < number; j++)
                    repeat += temp;
                for (let c = 0; c < repeat.length; c++)
                    st.push(repeat);
            }

            // if s[i] is not ']', simply push s[i] to the stack
            else
                st.push(s[i]);
        }
        let res = "";
        while (st.length > 0) {
            res =  st[st.length - 1] + res;
            st.pop();
        }
        return res;
    }

    let str = "3[b2[ca]]";
    document.write(decodeString(str));

// This code is contributed by divyesh072019.
</script>

Output

bcacabcacabcaca

Time Complexity: O(n)
Auxiliary Space: O(n)

Method 3: Without Stack

The given problem can be solved by traversing the encoded string character by character and maintaining a result string. Whenever a closing bracket is encountered, we can extract the substring enclosed within the corresponding opening bracket, and the number of times it needs to be repeated, and append the resulting string to the current result. We can continue this process until we reach the end of the input string.

Algorithm for better understanding.

1. Initialize an empty string to store the decoded output.
2. Traverse the input string character by character.
3. If the current character is not a closing bracket ‘]’, add it to the output string.
4. If the current character is a closing bracket, extract the substring enclosed within the corresponding opening bracket ‘[…]’, and the number of times it needs to be repeated say ‘num’.
5. Append the resulting string to the output string num times.
6. Repeat steps 3-5 until we reach the end of the input string.
7. Return the decoded string.

Below is the implementation of the above idea.

C++

// C++ implementation to decode string without stack
#include <algorithm>
#include <iostream>
#include <string>
using namespace std;

// Function to decode given encoded string
string decodeString(string s)
{

    // Declare a string variable to store the decoded
    // string.
    string result = "";

    // Traverse the encoded string character by character.
    for (int i = 0; i < s.length(); i++) {

        // If the current character is not a closing
        // bracket, append it to the result string.
        if (s[i] != ']') {
            result.push_back(s[i]);
        }

        // If the current character is a closing bracket
        else {

            // Create a temporary string to store the
            // substring within the corresponding opening
            // bracket.
            string temp = "";
            while (!result.empty()
                   && result.back() != '[') {
                temp.push_back(result.back());
                result.pop_back();
            }

            // Reverse the temporary string to obtain the
            // correct substring.
            reverse(temp.begin(), temp.end());

            // Remove the opening bracket from the result
            // string.
            result.pop_back();

            // Extract the preceding number and convert it
            // to an integer.
            string num = "";
            while (!result.empty() && result.back() >= '0'
                   && result.back() <= '9') {
                num.push_back(result.back());
                result.pop_back();
            }
            reverse(num.begin(), num.end());
            int int_num = stoi(num);

            // Append the substring to the result string,
            // repeat it to the required number of times.
            while (int_num--) {
                result += temp;
            }
        }
    }

    // Return the decoded string.
    return result;
}

// driver code
int main()
{
    string str = "3[b2[ca]]";
    cout << decodeString(str);
    return 0;
}

Output

bcacabcacabcaca

Time Complexity: O(n), where n is the length of the input string, as we traverse the input string character by character only once.
Auxiliary Space: O(n), as we are creating a temporary string and a number string.