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# Minimize bit flips so that each continuous segment of the same bit has an even length

Given a binary string S with an even size containing only 0 seconds and 1s, the task is to find the minimum flips (i.e. 0 to 1 or 1 to 0) such that each contiguous subsegment containing the same bits is an even size.

Example:

Enter: S = “1110011000”
Output: 3
Explanation: Change SS and S to ‘0’.
After that S becomes “110000000000”, can be divided into
“11” and “00000000”, which have lengths of 2 and 8, respectively.
There are other ways to operate 3 times like
“1111110000”, “1100001100”, “1111001100”.

Enter: 100110
Output: 3
Explanation: The given string can be converted to 000000
with 3 operations or 111111.

Approach: To solve the problem follow the observations below:

It can be said that we have to split the string into many contiguous binaries of length 2 where the string will be either 00 or 11. So we only need to take care of the order in which s[i] != s[i+1]

• Now to find the minimum order of operations like “1011” must be changed to “1111” and “0100” must be changed to “0000”.
• So, the minimum number of operations is 1. Therefore, after we make s[i] = s[i + 1] we will move i to i+2 to check if they are equal or not.

Follow the steps mentioned below to implement the idea:

• Repeat from i = 0 to N-1:
• Check if S[i] together with S[i+1].
• If they are the same then there is no need for a slight flip.
• If not, increase the number of flips by 1.
• After each check moved from I to me+2.
• The total number of flips counted is the minimum answer required.

Below is the implementation of the above approach:

## C++14

 ` ` `#include ` `using` `namespace` `std;` ` ` `int` `solve(string s)` `{` `    ` `    ``int` `ans = 0;` ` ` `    ``int` `n = s.size();` ` ` `    ` `    ``for` `(``int` `i = 0; i < n; i += 2) {` `        ``if` `(s[i] != s[i + 1])` `            ``ans++;` `    ``}` ` ` `    ` `    ``return` `ans;` `}` ` ` `int` `main()` `{` `    ``string S = ``"1110011000"``;` ` ` `    ` `    ``cout << solve(S);` `    ``return` `0;` `}`

Time complexity: ON)
Help Room: O(1)

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