### PROBLEM LINK:

**Author:** Praveen Dhinwa

**Tester:** Jingbo Shang

**Editorialist:** Utkarsh Saxena

### PROBLEM

Given a binary string of length 8. Make the string circular.

Count number of places where adjacent bits are different. Print “Uniform” if this count \le 2

### EXPLANATION

Since this problem was a cakewalk, it is quite straightforward to code.

There is quite less to explain apart from giving some observations.

##Bruteforce C++

```
for(int i=0;i<8;++i)
count += s* != s[(i+1)&7];
```

##Bruteforce Python

```
for i in range(8):
count += s* != s[i-1]
```

##Random observations

To have count=0 the string must have 8\space 0's or 8\space 1's.

It is not possible to have count=1.

To have count = 2, the string must have exactly one 1 or exactly one 0.

So for this problem the total number of 1 in the string can be 0, 1, 7, 8.

### AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.

Tester’s solution can be found here.