# 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: 3
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: 3
Explanation:
Initial String – “00000”
Flip the 2nd bit – “01111”
Flip the 3rd bit – “01000”
Flip the 5th bit – “01001”
Total Flips = 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program for the above approach ` `#include ` `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 `

## Java

 `// 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)); ` `    ``} ` `} `

## C#

 `// 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 `

Output:

```2
```

Time Complexity: O(N)

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Improved By : 29AjayKumar, rock_cool

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