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# 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```
Recommended Practice

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` `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 integerstack = ``new` `Stack<>();` `        ``Stack 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

 ``

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 ` `#include ` `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 st = ``new` `Vector();` `        `  `        ``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

 ``

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 ` `#include ` `#include ` `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.

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