Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).
You are given a target value to search. If found in the array return true, otherwise return false.
Example 1:
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true
Example 2:
Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false
Follow up:
This is a follow up problem to , where nums may contain duplicates.
Would this affect the run-time complexity? How and why?
// Binary Search
bool search(vector<int>& nums, int target) { // time: O(logn); space: O(1)
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return true;
if (nums[left] < nums[mid]) { // 1st half is in order
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else if (nums[left] > nums[mid]) { // 2nd half is in order
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
} else { // handle the case like: [3, 1, 2, 3, 3, 3]
++left;
}
}
return false;
}
class Solution:
def search(self, nums: List[int], target: int) -> bool:
left = 0
right = len(nums) - 1
while left <= right:
mid = left + (right - left) // 2
if nums[mid] == target:
return True
if nums[left] < nums[mid]:
if nums[left] <= target and target < nums[mid]:
right = mid - 1
else:
left = mid + 1
elif nums[left] > nums[mid]:
if nums[mid] < target and target <= nums[right]:
left = mid + 1
else:
right = mid - 1
else:
left = left + 1
return False