# Minimum flips to remove any consecutive 3 0s or 1s in given Binary string

• Difficulty Level : Medium
• Last Updated : 06 Sep, 2021

Given a binary string S consisting of N characters, the task is to find the minimum number of flips required such that there don’t exist three consecutive same characters.

Examples:

Input: S = “1100011”
Output: 1
Explanation:
Flip the character at index 3 modifies the string S “1101011” that have no three consecutive same characters. Therefore, the minimum number of flips required is 1.

Input: S = “0001111101”
Output: 2

Approach: The given problem can be solved by considering every three consecutive characters and if they are the same, then increase the count of flips required as one of the three characters is needed to be flipped. Follow the steps below to solve the problem:

• Initialize the variable, say count as 0 that stores the minimum number of flips required.
• If the size of the string is less than equal to 2, then return 0 as there is no need for any flips.
• Iterate over the range [0, N – 2) using the variable i and perform the following steps:
• If the character at indices i, (i + 1), and (i + 2) characters are the same, then increment the value of count by 1 and the value of i by 3.
• Otherwise, increment the value of i by 1.
• After performing the above steps, print the value of count as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the minimum number``// of flips to make all three pairs of``// consecutive characters different``int` `minFlips(string str)``{``    ``// Stores resultant count of pairs``    ``int` `count = 0;` `    ``// Base Case``    ``if` `(str.size() <= 2) {``        ``return` `0;``    ``}` `    ``// Iterate over the range [0, N - 2]``    ``for` `(``int` `i = 0; i < str.size() - 2;) {` `        ``// If the consecutive 3 numbers``        ``// are the same then increment``        ``// the count and the counter``        ``if` `(str[i] == str[i + 1]``            ``&& str[i + 2] == str[i + 1]) {``            ``i = i + 3;``            ``count++;``        ``}``        ``else` `{``            ``i++;``        ``}``    ``}` `    ``// Return the answer``    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``string S = ``"0011101"``;``    ``cout << minFlips(S);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``class` `GFG``{` `// Function to find the minimum number``// of flips to make all three pairs of``// consecutive characters different``static` `int` `minFlips(String str)``{``    ``// Stores resultant count of pairs``    ``int` `count = ``0``;``  ` `    ``// Base Case``    ``if` `(str.length() <= ``2``) {``        ``return` `0``;``    ``}``  ` `    ``// Iterate over the range [0, N - 2]``    ``for` `(``int` `i = ``0``; i < str.length() - ``2``ðŸ˜‰ {``  ` `        ``// If the consecutive 3 numbers``        ``// are the same then increment``        ``// the count and the counter``        ``if` `(str.charAt(i) == str.charAt(i+``1``)``            ``&& str.charAt(i+``2``) == str.charAt(i+``1``)) {``            ``i = i + ``3``;``            ``count++;``        ``}``        ``else` `{``            ``i++;``        ``}``    ``}``  ` `    ``// Return the answer``    ``return` `count;``}``  ` `// Driver Code``public` `static` `void` `main(String[] args)``{``     ``String S = ``"0011101"``;``   ``System.out.println(minFlips(S));``}``}` `// This code is contributed by dwivediyash`

## Python3

 `# python 3 program for the above approach``#``# Function to find the minimum number``# of flips to make all three pairs of``# consecutive characters different``def` `minFlips(st):` `    ``# Stores resultant count of pairs``    ``count ``=` `0` `    ``# Base Case``    ``if` `(``len``(st) <``=` `2``):``        ``return` `0` `    ``# Iterate over the range [0, N - 2]``    ``for` `i ``in` `range``(``len``(st) ``-` `2``):` `        ``# If the consecutive 3 numbers``        ``# are the same then increment``        ``# the count and the counter``        ``if` `(st[i] ``=``=` `st[i ``+` `1``]``                ``and` `st[i ``+` `2``] ``=``=` `st[i ``+` `1``]):``            ``i ``=` `i ``+` `3``            ``count ``+``=` `1` `        ``else``:``            ``i ``+``=` `1` `    ``# Return the answer``    ``return` `count` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``S ``=` `"0011101"``    ``print``(minFlips(S))` `    ``# This code is contributed by ukasp.`

## Javascript

 `  ```

## C#

 `// C# program for the above approach``using` `System;`  `public` `class` `GFG``{` `// Function to find the minimum number``// of flips to make all three pairs of``// consecutive characters different``static` `int` `minFlips(``string` `str)``{``    ``// Stores resultant count of pairs``    ``int` `count = 0;``  ` `    ``// Base Case``    ``if` `(str.Length <= 2) {``        ``return` `0;``    ``}``  ` `    ``// Iterate over the range [0, N - 2]``    ``for` `(``int` `i = 0; i < str.Length - 2;) {``  ` `        ``// If the consecutive 3 numbers``        ``// are the same then increment``        ``// the count and the counter``        ``if` `(str[i] == str[i+1]``            ``&& str[i+2] == str[i+1]) {``            ``i = i + 3;``            ``count++;``        ``}``        ``else` `{``            ``i++;``        ``}``    ``}``  ` `    ``// Return the answer``    ``return` `count;``}``  ` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``     ``string` `S = ``"0011101"``;``     ``Console.WriteLine(minFlips(S));``}``}` `// This code is contributed by AnkThon`

Output:

`1`

Time Complexity: O(N)
Auxiliary Space: O(1)

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