Maximize the decimal equivalent by flipping only a contiguous set of 0s

Given a binary number in the form of a string, the task is to print a binary equivalent obtained by flipping only one contiguous set of 0s such that the decimal equivalent of this binary number is maximum.
Note: Do not assume any trailing zeroes in the start of the binary number i.e. “0101” is given as “101”.
Examples: 

Input: s = “10101” 
Output: 11101 
Explanation: 
Here we can only flip the 2nd character of the string “10101” ( = 21) that will change it to “11101” (= 29). Since we are allowed to flip a continuous subarray, any more flipping will lead to decrease in decimal equivalent.
Input: s = “1000” 
Output: 1111 
Explanation: 
If we flip the continuous characters starting from position 1 till 3 we will get 1111 which is the maximum number possible in 4 bits i.e. 15. 

Approach: To solve the problem mentioned above we know that we have to increase the value of the binary equivalent. Therefore, we must increase the number of 1’s at higher position. Clearly, we can increase the value of the number by only flipping the initially occurring zeroes. Traverse the string and flip the first occurrence of zeroes until a 1 occurs, in this case, the loop must break. Print the resultant string.
Below is the implementation of the above approach:
 

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation to Maximize the value of
// the decimal equivalent given in the binary form
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the binary number
void flip(string& s)
{
    for (int i = 0; i < s.length(); i++) {
 
        // Check if the current number is 0
        if (s[i] == '0') {
 
            // Find the continuous 0s
            while (s[i] == '0') {
 
                // Replace initially
                // occurring 0 with 1
                s[i] = '1';
                i++;
            }
 
            // Break out of loop if 1 occurs
            break;
        }
    }
}
 
// Driver code
int main()
{
    string s = "100010001";
    flip(s);
 
    cout << s;
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation to maximize the value of
// the decimal equivalent given in the binary form
import java.util.*;
 
class GFG{
 
// Function to print the binary number
static void flip(String s)
{
    StringBuilder sb = new StringBuilder(s);
    for(int i = 0; i < sb.length(); i++)
    {
        
       // Check if the current number is 0
       if (sb.charAt(i) == '0')
       {
            
           // Find the continuous 0s
           while (sb.charAt(i) == '0')
           {
                
               // Replace initially
               // occurring 0 with 1
               sb.setCharAt(i, '1');
               i++;
           }
            
           // Break out of loop if 1 occurs
           break;
       }
    }
    System.out.println(sb.toString());
}
 
// Driver code
public static void main(String[] args)
{
    String s = "100010001";
    flip(s);
}
}
 
// This code is contributed by offbeat
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation to
# Maximize the value of the
# decimal equivalent given
# in the binary form
 
# Function to print the binary
# number
def flip(s):
    s = list(s)
    for i in range(len(s)):
 
        # Check if the current number
        # is 0
        if(s[i] == '0'):
 
            # Find the continuous 0s
            while(s[i] == '0'):
 
                # Replace initially
                # occurring 0 with 1
                s[i] = '1'
                i += 1
            s = ''.join(map(str, s))
 
            # return the string and
            # break the loop
            return s
 
# Driver code
s = "100010001"
print(flip(s))
 
# This code is contributed by avanitrachhadiya2155
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation to maximize the value of
// the decimal equivalent given in the binary form
using System;
 
class GFG{
 
// Function to print the binary number
static String flip(char []s)
{
    for(int i = 0; i < s.Length; i++)
    {
        
       // Check if the current number is 0
       if (s[i] == '0')
       {
            
           // Find the continuous 0s
           while (s[i] == '0')
           {
                
               // Replace initially
               // occurring 0 with 1
               s[i] = '1';
               i++;
           }
            
           // Break out of loop if 1 occurs
           break;
       }
    }
    return new String(s);
}
 
// Driver code
public static void Main(String[] args)
{
    String s = "100010001";
     
    Console.WriteLine(flip(s.ToCharArray()));
}
}
 
// This code is contributed by Rohit_ranjan
chevron_right

Output: 
111110001

 





Student of Software engineering in Delhi Technological University My tech interests are Machine learning , software testing and artificial intelligence

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :