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# Minimum number of steps needed to remove the substring K from given string

Given a binary string S, and a substring K, the task is to find the minimum no of steps required to flip the characters in a binary string such that it doesn’t contain the given substring K. Note: In one step we can change 0 to 1 or vice versa.
Examples:

Input: S = “0111011”, K = “011”
Output:
Explanation:
In the above string we have the substring 011 for 2 times, hence change the 3rd bit and 7th bit.

Input: S = “010010”, K = “0100”
Output:
Explanation:
In the above string we have the substring 0100 for 1 time, change the 4th bit of the string to 1.

Naive Approach: The naive approach is to use the Pattern Searching. Iterate through the string for N! times(N = length of the binary string). In every iteration check for the substring K and if its a match then increment the count. At last, print the count which will be the number of steps required to make the string such that it doesn’t contain the given substring K.

Time Complexity: O(M*N) where M is the length of substring and N is the length of the binary string.
Auxiliary Space: O(1)

Efficient Approach: To optimize the above method, the idea is to replace the substring with an empty string and subtract the resultant string length from original string length. After this divide the resultant string with the length of the given substring K to get the minimum no of steps required to flip the characters of the given string S so that it doesn’t contain the given substring K.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include``#include ``using` `namespace` `std;` `// Function that counts the total``// number of character to be flipped``// such the string b doesn't contains``// string sub as a substring``int` `flipCharacters(string b,``                   ``string sub)``{``    ` `    ``// Replacing the substring with ""``    ``// and find the difference between``    ``// the length of the resultant``    ``// string and the original``    ``// string length``    ``int` `res = b.size();``    ``boost::replace_all(b, sub, ``""``);``    ``res = res - b.size();``             ` `    ``// Divide the result``    ``// with substring length``    ``int` `temp = res / sub.size();` `    ``// Return the final count``    ``return` `temp;``}` `// Driver Code``int` `main()``{``    ` `    ``// Given string S and substring K``    ``string S = ``"010010"``;``    ``string K = ``"0100"``;` `    ``// Function call``    ``int` `result = flipCharacters(S, K);` `    ``// Print the minimum flip``    ``cout << (result);``}` `// This code is contributed by Amit Katiyar`

## Java

 `// Java program for the above approach``import` `java.util.*;` `public` `class` `Test {` `    ``// Function that counts the total``    ``// number of character to be flipped``    ``// such the string b doesn't contains``    ``// string sub as a substring``    ``static` `int` `flipCharacters(String b,``                              ``String sub)``    ``{` `        ``// Replacing the substring with ""``        ``// and find the difference between``        ``// the length of the resultant``        ``// string and the original``        ``// string length` `        ``int` `res = b.length()``                  ``- b.replaceAll(sub, ``""``)``                        ``.length();` `        ``// Divide the result``        ``// with substring length` `        ``int` `temp = res / sub.length();` `        ``// Return the final count``        ``return` `temp;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``// Given string S and substring K``        ``String S = ``"010010"``;``        ``String K = ``"0100"``;` `        ``// Function Call``        ``int` `result = flipCharacters(S, K);` `        ``// Print the minimum flip``        ``System.out.println(result);``    ``}``}`

## Python3

 `# Python3 program for the above approach``def` `flipCharacters(b, sub):``    ` `    ``# Replace the substring with``    ``# "" (emptryString)``    ``b1 ``=` `b.replace(sub, "")``    ` `    ``n ``=` `int``((``len``(b)``-``len``(b1))``/``len``(sub))``    ` `    ``return` `n` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:` `# Given string S and substring K``    ``S ``=``"010010"``    ``X ``=``"0100"` `# Function Call``    ``result ``=` `flipCharacters(S, X)` `# Print the minimum flip``    ``print``(result)`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG``{`` ` `  ``// Function that counts the total``  ``// number of character to be flipped``  ``// such the string b doesn't contains``  ``// string sub as a substring``  ``static` `int` `flipCharacters(``string` `b,``                            ``string` `sub)``  ``{` `    ``// Replacing the substring with ""``    ``// and find the difference between``    ``// the length of the resultant``    ``// string and the original``    ``// string length` `    ``int` `res = b.Length -``              ``b.Replace(sub, ``""``).Length;` `    ``// Divide the result``    ``// with substring length``    ``int` `temp = res / sub.Length;` `    ``// Return the final count``    ``return` `temp;``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args)``  ``{` `    ``// Given string S and substring K``    ``string` `S = ``"010010"``;``    ``string` `K = ``"0100"``;` `    ``// Function Call``    ``int` `result = flipCharacters(S, K);` `    ``// Print the minimum flip``    ``Console.Write(result);``  ``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``

Output:

`1`

Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(1)