Search in Rotated Sorted Array
Given a rotated sorted array nums (with distinct values) and a target, return the index of target or -1 if not found.
You must achieve O(log n) time complexity.
Example:
4,5,6,7,0,1,2 0
4
- The given array is rotated, so we need to find the pivot point where the rotation occurred.
- We use a modified binary search algorithm to achieve O(log n) time complexity, dividing the search space in half at each step.
- The target value 0 is less than the middle element of the array, so we repeat the search in the right half of the array: [0,1,2].
- Since 0 is found at the first position of the right half, which is the 4th index in the original array (using 0-based indexing), the function returns 4.
Constraints:
- 1 <= len(nums) <= 5000
- -10^4 <= nums[i] <= 10^4
- All values are unique
Background Knowledge
The problem "Search in Rotated Sorted Array" involves a sorted array that has been rotated (or shifted) by some number of positions. This means that the array was initially sorted in ascending order, but then its elements were rotated to the right (or left) by a certain number of steps. For example, the array [1, 2, 3, 4, 5, 6, 7] rotated by 3 steps to the right becomes [5, 6, 7, 1, 2, 3, 4]. To solve this problem efficiently, we need to understand how to search in a sorted array and how to adapt this search to a rotated sorted array.
The key concept here is binary search, which is an efficient algorithm for finding an item from a sorted list of items. It works by repeatedly dividing in half the portion of the list that could contain the item, until you've narrowed the possible locations to just one. In the context of a rotated sorted array, we need to modify the binary search algorithm to account for the rotation. This involves identifying which half of the array is still sorted and deciding which half to continue searching in based on the target value.
Understanding the properties of a rotated sorted array is crucial. We know that the array is divided into two halves: one that is sorted and one that is not. By comparing the middle element of the array to the first and last elements, we can determine which half is sorted. This insight allows us to apply a modified binary search strategy to find the target element efficiently.
Algorithm/Approach
The algorithm pattern to solve this type of problem is a variation of the binary search algorithm. The general approach involves:
- Identifying the middle element of the current search range
- Comparing the middle element to the target and adjusting the search range accordingly
- Determining which half of the array is sorted and using this information to guide the search
This approach takes advantage of the fact that even though the array is rotated, one half of it remains sorted. By cleverly choosing which half to search in at each step, we can maintain an efficient search process.
Step-by-Step Strategy
To implement the solution:
- Initialize the search range to the entire array.
- Calculate the middle index of the current search range.
- Compare the middle element to the target value.
- Determine which half of the array is sorted by comparing the middle element to the first and last elements of the search range.
- Based on the target value and the sorted half, decide which half to continue searching in and adjust the search range accordingly.
- Repeat steps 2-5 until the target is found or the search range becomes empty.
Common Pitfalls
When implementing, watch out for:
- Incorrectly determining which half of the array is sorted
- Failing to adjust the search range correctly based on the target value and the sorted half
- Not handling edge cases, such as an empty array or an array with a single element
Time & Space Complexity
The expected time complexity for this problem is O(logn), where n is the number of elements in the array. This is because we are using a modified binary search algorithm that divides the search space in half at each step. The space complexity is O(1), as we only need a constant amount of space to store the search range and the target value.