81. Search in Rotated Sorted Array II
Problem¶
There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).
Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].
Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.
You must decrease the overall operation steps as much as possible.
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
Constraints:
1 <= nums.length <= 5000-104 <= nums[i] <= 104numsis guaranteed to be rotated at some pivot.-104 <= target <= 104
Follow up: This problem is similar to 33. Search in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?
Solve¶
Remove Duplicate + Find shift point + Binary search¶
O(n)
Some conclusion¶
While there is a lot of similarity from 33. Search in Rotated Sorted Array . The main breaking point is that we can’t effectively find the shift point as if nums[mid] == nums[0], it could either be left, or right because of duplication. Both example could be:
So to preventing this, we should at least remove duplication on either end, here is a snippet code that I try to clear the duplicate on nums array left side:
for i in range(n-1,-1,-1):
if nums[i] == nums[0]:
continue
self.nums = nums[:i+1]
n = len(self.nums)
break
if n == 0:
return False
Final code should be the same as 33. Search in Rotated Sorted Array
class Solution:
def findShiftPoint(self):
shiftPoint = -1
isFound = False
pointerLeftIndex = 0
pointerRightIndex = len(self.nums)
while True:
midIndex = (pointerLeftIndex + pointerRightIndex) // 2
if pointerLeftIndex == midIndex:
break
if self.nums[midIndex] >= self.nums[0]:
pointerLeftIndex = midIndex
else:
pointerRightIndex = midIndex
if pointerLeftIndex+1 == pointerRightIndex:
shiftPoint = len(self.nums)-1 - pointerLeftIndex
isFound = True
return (shiftPoint, isFound)
def binarySearch(self, startIndex,endIndex, target):
targetIndex, isFound = -1, False
left = startIndex-1
right = endIndex +1
while True:
midIndex = (left + right) // 2
if left == midIndex:
break
if self.nums[midIndex] <= target:
left = midIndex
else:
right = midIndex
if left+1 == right:
targetIndex = left
if startIndex <= targetIndex <= endIndex:
isFound = self.nums[targetIndex] == target
return (targetIndex, isFound)
def search(self, nums: List[int], target: int) -> int:
self.nums = nums
n = len(nums)
self.target = target
for i in range(n-1,-1,-1):
if nums[i] == nums[0]:
continue
self.nums = nums[:i+1]
n = len(self.nums)
break
if n == 0:
return False
result = -1
shiftPoint, isFound = self.findShiftPoint()
targetIndex, isFound = self.binarySearch(0, n- shiftPoint -1, target)
if isFound:
return True
targetIndex, isFound = self.binarySearch(n- shiftPoint,n-1, target)
if isFound:
return True
return False
Time complexity: Worst case O(n)
- As we need to remove duplicate on left/right side, which in theory can be the whole array, costing O(n) time
- Other binary finding cost O(log n)
Created : August 16, 2023