Given a binary string S , the task is to find the smallest string possible by removing all occurrences of substrings “01” and “11”. After removal of any substring, concatenate the remaining parts of the string.
Examples:
Input: S = “0010110”
Output:
Length = 1 String = 0
Explanation: String can be transformed by the following steps:
0010110 → 00110 → 010 → 0.
Since no occurrence of substrings 01 and 11 are remaining, the string “0” is of minimum possible length 1.Input: S = “0011101111”
Output: Length = 0
Explanation:
String can be transformed by the following steps:
0011101111 → 01101111 → 011011 → 1011 → 11 → “”.
Approach: To solve the problem, the idea is to observe the following cases:
- 01 and 11 mean that ?1 can be removed where ‘?’ can be 1 or 0.
- The final string will always be in the form 1000… or 000…
This problem can be solved by maintaining a Stack while processing the given string S from left to right. If the current binary digit is 0, add it to the stack, if the current binary digit is 1, remove the top bit from the stack. If the stack is empty, then push the current bit to the stack. Follow the below steps to solve the problem:
- Initialize a Stack to store the minimum possible string.
- Traverse the given string over the range [0, N – 1].
- If the stack is empty, push the current binary digit S[i] in the stack.
- If the stack is not empty and the current bit S[i] is 1 then remove the top bit from the stack.
- If the current element S[i] is 0 then, push it to the stack.
- Finally, append all the elements present in the stack from top to bottom and print it as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find minimum // length of the given string void findMinLength(string s, int n) { // Initialize a stack stack< int > st; // Traverse the string for ( int i = 0; i < n; i++) { // If the stack is empty if (st.empty()) // Push the character st.push(s[i]); // If the character is 1 else if (s[i] == '1' ) // Pop the top element st.pop(); // Otherwise else // Push the character // to the stack st.push(s[i]); } // Initialize length int ans = 0; // Append the characters // from top to bottom vector< char > finalStr; // Until Stack is empty while (!st.empty()) { ans++; finalStr.push_back(st.top()); st.pop(); } // Print the final string size cout << "Length = " << ans; // If length of the string is not 0 if (ans != 0) { // Print the string cout << "\nString = " ; for ( int i = 0; i < ans; i++) cout << finalStr[i]; } } // Driver Code int main() { // Given string string S = "101010" ; // String length int N = S.size(); // Function call findMinLength(S, N); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Function to find minimum // length of the given string static void findMinLength(String s, int n) { // Initialize a stack Stack<Character> st = new Stack<>(); // Traverse the string for ( int i = 0 ; i < n; i++) { // If the stack is empty if (st.empty()) // Push the character st.push(s.charAt(i)); // If the character is 1 else if (s.charAt(i) == '1' ) // Pop the top element st.pop(); // Otherwise else // Push the character // to the stack st.push(s.charAt(i)); } // Initialize length int ans = 0 ; // Append the characters // from top to bottom Vector<Character> finalStr = new Vector<Character>(); // Until Stack is empty while (st.size() > 0 ) { ans++; finalStr.add(st.peek()); st.pop(); } // Print the final string size System.out.println( "Length = " + ans); // If length of the string is not 0 if (ans != 0 ) { // Print the string System.out.print( "String = " ); for ( int i = 0 ; i < ans; i++) System.out.print(finalStr.get(i)); } } // Driver Code public static void main(String args[]) { // Given string String S = "101010" ; // String length int N = S.length(); // Function call findMinLength(S, N); } } // This code is contributed by SURENDRA_GANGWAR |
Python3
# Python3 program for the above approach from collections import deque # Function to find minimum length # of the given string def findMinLength(s, n): # Initialize a stack st = deque() # Traverse the string from # left to right for i in range (n): # If the stack is empty, # push the character if ( len (st) = = 0 ): st.append(s[i]) # If the character # is B, pop from stack elif (s[i] = = '1' ): st.pop() # Otherwise, push the # character to the stack else : st.append(s[i]) # Stores resultant string ans = 0 finalStr = [] while ( len (st) > 0 ): ans + = 1 finalStr.append(st[ - 1 ]); st.pop() # Print the final string size print ( "The final string size is: " , ans) # If length is not 0 if (ans = = 0 ): print ( "The final string is: EMPTY" ) # Print the string else : print ( "The final string is: " , * finalStr) # Driver Code if __name__ = = '__main__' : # Given string s = "0010110" # String length n = 7 # Function Call findMinLength(s, n) |
C#
// C# program for the above approach using System; using System.Collections.Generic; class GFG{ // Function to find minimum // length of the given string static void findMinLength(String s, int n) { // Initialize a stack Stack< char > st = new Stack< char >(); // Traverse the string for ( int i = 0; i < n; i++) { // If the stack is empty if (st.Count == 0) // Push the character st.Push(s[i]); // If the character is 1 else if (s[i] == '1' ) // Pop the top element st.Pop(); // Otherwise else // Push the character // to the stack st.Push(s[i]); } // Initialize length int ans = 0; // Append the characters // from top to bottom List< char > finalStr = new List< char >(); // Until Stack is empty while (st.Count > 0) { ans++; finalStr.Add(st.Peek()); st.Pop(); } // Print the readonly string size Console.WriteLine( "Length = " + ans); // If length of the string is not 0 if (ans != 0) { // Print the string Console.Write( "String = " ); for ( int i = 0; i < ans; i++) Console.Write(finalStr[i]); } } // Driver Code public static void Main(String []args) { // Given string String S = "101010" ; // String length int N = S.Length; // Function call findMinLength(S, N); } } // This code is contributed by Amit Katiyar |
Length = 2 String = 01
Time Complexity: O(N)
Auxiliary Space: O(N)
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