Minimum flips required to form given binary string where every flip changes all bits to its right as well

Given a string S, the task is to find minimum flips required to convert an initial binary string consisting of only zeroes to S where every flip of a character flips all succeeding characters as well. 
Examples: 
 

Input: S = “01011” 
Output:
Explanation: 
Initial String – “00000” 
Flip the 2nd bit – “01111” 
Flip the 3rd bit – “01000” 
Flip the 4th bit – “01011” 
Total Flips = 3
Input: S = “01001” 
Output:
Explanation: 
Initial String – “00000” 
Flip the 2nd bit – “01111” 
Flip the 3rd bit – “01000” 
Flip the 5th bit – “01001” 
Total Flips = 3 
 

 

Approach: 
In order to solve the problem, follow the steps below: 
 

  • Store ‘1’ in curr initially.
  • Traverse S and find the first occurrence of curr. Increase count when curr is encountered. Store ‘0’ if curr is ‘1’ or vice versa.
  • Repeat the above step for entire traversal of S.
  • Final value of count gives the required answer.

Below is the implementation of the above approach.



C++

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// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
  
// Function to return the count
// of minimum flips required
int minFlips(string target)
{
    char curr = '1';
    int count = 0;
    for(int i = 0; i < target.length(); i++)
    {
          
       // If curr occurs in the final string
       if (target[i] == curr)
       {
           count++;
             
           // Switch curr to '0' if '1'
           // or vice-versa
           curr = (char)(48 + (curr + 1) % 2);
       }
    }
    return count;
}
  
// Driver Code
int main()
{
    string S = "011000";
      
    cout << (minFlips(S));
}
  
// This code is contributed by rock_cool

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Java

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// Java program for the above approach
import java.util.Arrays;
  
public class GFG {
    // Function to return the count of
    // minimum flips required
    public static int minFlips(String target)
    {
  
        char curr = '1';
        int count = 0;
        for (int i = 0; i < target.length(); i++) {
  
            // If curr occurs in the final string
            if (target.charAt(i) == curr) {
  
                count++;
  
                // Switch curr to '0' if '1'
                // or vice-versa
                curr = (char)(48 + (curr + 1) % 2);
            }
        }
  
        return count;
    }
  
    // Driver Code
    public static void main(String args[])
    {
  
        String S = "011000";
        System.out.println(minFlips(S));
    }
}

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Python3

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# Python3 program for the above approach
  
# Function to return the count
# of minimum flips required
def minFlips(target):
  
    curr = '1'
    count = 0
      
    for i in range(len(target)):
          
        # If curr occurs in the final string
        if (target[i] == curr):
            count += 1
              
            # Switch curr to '0' if '1'
            # or vice-versa
            curr = chr(48 + (ord(curr) + 1) % 2)
      
    return count
  
# Driver Code
if __name__ == "__main__":
      
    S = "011000"
      
    print(minFlips(S))
  
# This code is contributed by chitranayal

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C#

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// C# program for the above approach
using System;
  
class GFG{
      
// Function to return the count of
// minimum flips required
public static int minFlips(String target)
{
    char curr = '1';
    int count = 0;
    for(int i = 0; i < target.Length; i++) 
    {
          
       // If curr occurs in the readonly string
       if (target[i] == curr)
       {
           count++;
             
           // Switch curr to '0' if '1'
           // or vice-versa
           curr = (char)(48 + (curr + 1) % 2);
       }
    }
    return count;
}
  
// Driver code
public static void Main(String []args)
{
    String S = "011000";
    Console.WriteLine(minFlips(S));
}
}
  
// This code is contributed by 29AjayKumar

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Output: 

2

Time Complexity: O(N)
 

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