LC81. Search in Rotated Sorted Array II¶

Problem Description¶

LeetCode Problem 81: 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.

Clarification¶

sorted and rotated
have duplicates

Assumption¶

Solution¶

Approach - Binary Search¶

This is a follow-up problem to LC33 - Search in Rotated Sorted Array. The difference is that it contains duplicates. We can still divide the array into a sorted part and a unsorted part and then search. Due to duplicates,

For situation 1 in LC33, nums[left] <= nums[mid] <= nums[right], when nums[left] == nums[mid] == nums[right], we don’t know which part (left or right) is sorted. For example, [3,1,3,3,3,3,3] with left part unsorted and [3, 3, 3, 3, 3, 1, 3] with right part unsorted. For this case, if target != nums[mid], we can increase left and decrease right by 1.

Python

class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        left, right = 0, len(nums) - 1

        while left <= right:
            mid = (left + right) // 2
            if nums[mid] == target:
                return True
            elif nums[left] == nums[mid] and nums[mid] == nums[right]:
                left += 1
                right -= 1
            elif nums[left] <= nums[mid]:  # (1)
                # [left, mid] non-decreasing
                if nums[left] <= target and target < nums[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            else:
                # [mid, right] non-decreasing
                if nums[mid] < target and target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1

        return False

Add = to handle case `left == mid``

Complexity Analysis¶

Time complexity: \(O(\log n)\) for the average, \(O(n)\) for the worst case
In the worst case, just move left pointer by one, which needs n steps. So the time complexity is \(O(n)\). For the average case, since using binary search, the time complexity is \(O(\log n)\).
Space complexity: \(O(1)\)
Only use limited variables for indices.

Test¶

Only two numbers, e.g. [3, 1]
Test array with `nums[left] == nums[mid]``