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++ 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 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 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# 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 |
<script> // JavaScript program for the above approach // Function to find minimum // length of the given string function findMinLength(s, n)
{ // Initialize a stack
var st = [];
// Traverse the string
for ( var i = 0; i < n; i++) {
// If the stack is empty
if (st.length==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
var ans = 0;
// Append the characters
// from top to bottom
var finalStr = [];
// Until Stack is empty
while (st.length!=0) {
ans++;
finalStr.push(st[st.length-1]);
st.pop();
}
// Print the final string size
document.write( "Length = " + ans);
// If length of the string is not 0
if (ans != 0) {
// Print the string
document.write( "<br>String = " );
for ( var i = 0; i < ans; i++)
document.write( finalStr[i]);
}
} // Driver Code // Given string var S = "101010" ;
// String length var N = S.length;
// Function call findMinLength(S, N); </script> |
Length = 2 String = 01
Time Complexity: O(N)
Auxiliary Space: O(N)