# Min flips of continuous characters to make all characters same in a string

Given a string consisting only of 1’s and 0’s. In one flip we can change any continuous sequence of this string. Find this minimum number of flips so the string consist of same characters only.
Examples:

`Input : 00011110001110Output : 2We need to convert 1's sequenceso string consist of all 0's.Input : 010101100011Output : 4`

Method 1 (Change in value encountered)
We need to find the min flips in string so all characters are equal. All we have to find numbers of sequence which consisting of 0’s or 1’s only. Then number of flips required will be half of this number as we can change all 0’s or all 1’s.

## C++

 `// CPP program to find min flips in binary` `// string to make all characters equal` `#include ` `using` `namespace` `std;`   `// To find min number of flips in binary string` `int` `findFlips(``char` `str[], ``int` `n)` `{` `    ``char` `last = ``' '``; ``int` `res = 0;`   `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// If last character is not equal` `        ``// to str[i] increase res` `        ``if` `(last != str[i])` `            ``res++;` `        ``last = str[i];` `    ``}`   `    ``// To return min flips` `    ``return` `res / 2;` `}`   `// Driver program to check findFlips()` `int` `main()` `{` `    ``char` `str[] = ``"00011110001110"``;` `    ``int` `n = ``strlen``(str);`   `    ``cout << findFlips(str, n);`   `    ``return` `0;` `}`

## Java

 `// Java program to find min flips in binary` `// string to make all characters equal` `public` `class` `minFlips {`   `    ``// To find min number of flips in binary string` `    ``static` `int` `findFlips(String str, ``int` `n)` `    ``{` `        ``char` `last = ``' '``; ``int` `res = ``0``;`   `        ``for` `(``int` `i = ``0``; i < n; i++) {`   `            ``// If last character is not equal` `            ``// to str[i] increase res` `            ``if` `(last != str.charAt(i))` `                ``res++;` `            ``last = str.charAt(i);` `        ``}`   `        ``// To return min flips` `        ``return` `res / ``2``;` `    ``}`   `    ``// Driver program to check findFlips()` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``String str = ``"00011110001110"``;` `        ``int` `n = str.length();`   `        ``System.out.println(findFlips(str, n));` `    ``}` `}`

## Python 3

 `# Python 3 program to find min flips in ` `# binary string to make all characters equal`   `# To find min number of flips in` `# binary string` `def` `findFlips(``str``, n):`   `    ``last ``=` `' '` `    ``res ``=` `0`   `    ``for` `i ``in` `range``( n) :`   `        ``# If last character is not equal` `        ``# to str[i] increase res` `        ``if` `(last !``=` `str``[i]):` `            ``res ``+``=` `1` `        ``last ``=` `str``[i]`   `    ``# To return min flips` `    ``return` `res ``/``/` `2`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    `  `    ``str` `=` `"00011110001110"` `    ``n ``=` `len``(``str``)`   `    ``print``(findFlips(``str``, n))`   `# This code is contributed by ita_c`

## C#

 `// C# program to find min flips in` `// binary string to make all` `// characters equal` `using` `System;`   `public` `class` `GFG {`   `    ``// To find min number of flips` `    ``// in binary string` `    ``static` `int` `findFlips(String str, ``int` `n)` `    ``{` `        ``char` `last = ``' '``; ``int` `res = 0;`   `        ``for` `(``int` `i = 0; i < n; i++) {`   `            ``// If last character is not` `            ``// equal to str[i] increase` `            ``// res` `            ``if` `(last != str[i])` `                ``res++;` `            ``last = str[i];` `        ``}`   `        ``// To return min flips` `        ``return` `res / 2;` `    ``}`   `    ``// Driver program to check findFlips()` `    ``public` `static` `void` `Main()` `    ``{` `        ``String str = ``"00011110001110"``;` `        ``int` `n = str.Length;`   `        ``Console.Write(findFlips(str, n));` `    ``}` `}`   `// This code is contributed by nitin mittal`

## Javascript

 ``

## PHP

 ``

Output

```2

```

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

Method 2(Count continuous 0 and 1 )

We can count the number of continuous 0 and continuous 1.
Since we have to take minimum, we use min function to take value with less continuous number. And output it.

Procedure:- Take two variable to count continuous 0 and 1. If 0 is encountered increment countZero and skip 0’s in continuation. Do same with 1 and its continuation. In the end the variables with less value is sent as output

## C++

 `#include ` `#include `   `using` `namespace` `std;`   `int` `main()` `{` `    ``string str = ``"010101100011"``;` `    ``int` `n = str.size();`   `    ``int` `countZero = 0;` `    ``int` `countOne = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(str[i] == ``'0'``) {` `            ``countZero++;` `            ``while` `((i + 1) != n && str[i + 1] == ``'0'``) {` `                ``i++;` `                ``// cout<<"Skip 0";` `            ``}` `        ``}` `        ``else` `{` `            ``countOne++;` `            ``while` `((i + 1) != n && str[i + 1] == ``'1'``) {` `                ``i++;` `                ``// cout<<"Skip 1";` `            ``}` `        ``}` `    ``}` `    ``int` `mini = min(countOne, countZero);` `    ``cout << mini << endl;`   `    ``return` `0;` `}`

## Java

 `public` `class` `Main {` `    ``public` `static` `void` `main(String[] args) {` `        ``String str = ``"010101100011"``;` `        ``int` `n = str.length();`   `        ``int` `countZero = ``0``;` `        ``int` `countOne = ``0``;` `        `  `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(str.charAt(i) == ``'0'``) {` `                ``countZero++;` `                ``while` `((i + ``1``) != n && str.charAt(i + ``1``) == ``'0'``) {` `                    ``i++;` `                    ``// System.out.println("Skip 0");` `                ``}` `            ``} ``else` `{` `                ``countOne++;` `                ``while` `((i + ``1``) != n && str.charAt(i + ``1``) == ``'1'``) {` `                    ``i++;` `                    ``// System.out.println("Skip 1");` `                ``}` `            ``}` `        ``}`   `        ``int` `mini = Math.min(countOne, countZero);` `        ``System.out.println(mini);` `    ``}` `}`

## Python3

 `str` `=` `"010101100011"` `n ``=` `len``(``str``)`   `countZero ``=` `0`  `# Counter for consecutive '0's` `countOne ``=` `0`   `# Counter for consecutive '1's` `i ``=` `0`          `# Index for string traversal`   `while` `i < n:` `    ``# Check for consecutive '0's` `    ``if` `str``[i] ``=``=` `'0'``:` `        ``countZero ``+``=` `1` `        ``# Count the consecutive '0's` `        ``while` `i ``+` `1` `!``=` `n ``and` `str``[i ``+` `1``] ``=``=` `'0'``:` `            ``i ``+``=` `1`  `# Move to the next character` `            ``# print("Skip 0")  # Optional: Uncomment to display skipping '0's`   `    ``# Check for consecutive '1's` `    ``else``:` `        ``countOne ``+``=` `1` `        ``# Count the consecutive '1's` `        ``while` `i ``+` `1` `!``=` `n ``and` `str``[i ``+` `1``] ``=``=` `'1'``:` `            ``i ``+``=` `1`  `# Move to the next character` `            ``# print("Skip 1")  # Optional: Uncomment to display skipping '1's`   `    ``i ``+``=` `1`  `# Move to the next character after consecutive count`   `# Calculate the minimum count between consecutive '0's and '1's` `mini ``=` `min``(countOne, countZero)`   `# Print the minimum count` `print``(mini)`

## C#

 `using` `System;`   `class` `Program` `{` `    ``static` `void` `Main()` `    ``{` `        ``string` `str = ``"010101100011"``;` `        ``int` `n = str.Length;`   `        ``int` `countZero = 0;` `        ``int` `countOne = 0;`   `        ``for` `(``int` `i = 0; i < n; i++)` `        ``{` `            ``if` `(str[i] == ``'0'``)` `            ``{` `                ``countZero++; ``// Increment the count of '0's` `                ``while` `((i + 1) != n && str[i + 1] == ``'0'``)` `                ``{` `                    ``i++; ``// Skip consecutive '0's` `                    ``// Console.WriteLine("Skip 0");` `                ``}` `            ``}` `            ``else` `            ``{` `                ``countOne++; ``// Increment the count of '1's` `                ``while` `((i + 1) != n && str[i + 1] == ``'1'``)` `                ``{` `                    ``i++; ``// Skip consecutive '1's` `                    ``// Console.WriteLine("Skip 1");` `                ``}` `            ``}` `        ``}`   `        ``int` `mini = Math.Min(countOne, countZero); ``// Find the minimum count of '0's and '1's` `        ``Console.WriteLine(mini); ``// Output the result` `    ``}` `}`

## Javascript

 `function` `countConsecutive(str) {` `    ``const n = str.length;`   `    ``let countZero = 0;` `    ``let countOne = 0;`   `    ``for` `(let i = 0; i < n; i++) {` `        ``if` `(str[i] === ``'0'``) {` `            ``countZero++;` `            ``while` `(i + 1 < n && str[i + 1] === ``'0'``) {` `                ``i++;` `                ``// console.log("Skip 0");` `            ``}` `        ``} ``else` `{` `            ``countOne++;` `            ``while` `(i + 1 < n && str[i +1] === ``'1'``) {` `                ``i++;` `                ``// console.log("Skip 1");` `            ``}` `        ``}` `    ``}`   `    ``const mini = Math.min(countOne, countZero);` `    ``console.log(mini);` `}`   `const str = ``"010101100011"``;` `countConsecutive(str);`

Output

```4

```

Time Complexity: O(n)

Auxiliary Space: O(1)

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