```
public boolean search(int[] nums, int target) {
int start = 0, end = nums.length - 1, mid = -1;
while(start <= end) {
mid = (start + end) / 2;
if (nums[mid] == target) {
return true;
}
//If we know for sure right side is sorted or left side is unsorted
if (nums[mid] < nums[end] || nums[mid] < nums[start]) {
if (target > nums[mid] && target <= nums[end]) {
start = mid + 1;
} else {
end = mid - 1;
}
//If we know for sure left side is sorted or right side is unsorted
} else if (nums[mid] > nums[start] || nums[mid] > nums[end]) {
if (target < nums[mid] && target >= nums[start]) {
end = mid - 1;
} else {
start = mid + 1;
}
//If we get here, that means nums[start] == nums[mid] == nums[end], then shifting out
//any of the two sides won't change the result but can help remove duplicate from
//consideration, here we just use end-- but left++ works too
} else {
end--;
}
}
return false;
}
```

In case anyone wonders, yes I agree that we don't need to check two parts. It's just that Doing that can slightly boost the performance, no asymptotic difference though.