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:
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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
Method 1(Using 2 stacks)
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 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 Programint main(){ string str = "3[b2[ca]]"; cout << decode(str) << endl; return 0;} |
Java
// Java program to decode a string recursively// encoded as count followed substringimport 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 codeif __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 substringusing 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 Codepublic 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
<!—-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 codeint main(){ string str = "3[b2[ca]]"; cout << decodeString(str); return 0;} |
Output
bcacabcacabcaca
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