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
- nums was rotated at some pivot
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.
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