search in rotated sorted array ii leetcode 81 python
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certainly! the problem "search in rotated sorted array ii" (leetcode 81) is a variation of searching in a rotated sorted array, where duplicates are allowed.
problem statement
you are given an integer array `nums` sorted in non-decreasing order, possibly rotated at some pivot unknown to you beforehand. you need to determine if a target value exists in the array. the array may contain duplicates.
key points to consider
1. **rotated sorted array**: the array was originally sorted but may have been rotated at some pivot.
2. **duplicates**: since the array can contain duplicates, this affects our search strategy.
3. **goal**: return `true` if the target is found, otherwise return `false`.
approach
1. **binary search**: we can use a modified binary search approach to solve this problem efficiently. the time complexity can be reduced to \(o(\log n)\) in the average case, but in the worst case with duplicates, it can degrade to \(o(n)\).
2. **handling duplicates**: when we encounter duplicates, we cannot decide which side of the array is sorted. in such cases, we can incrementally move the left pointer to skip duplicates.
steps
1. initialize two pointers, `left` and `right`, to the start and end of the array.
2. calculate the mid-point.
3. compare the mid-point value with the target:
if they are equal, return `true`.
determine which side of the array is sorted:
if the left side is sorted:
check if the target is between the left and mid-point values.
if the right side is sorted:
check if the target is between the mid-point and right values.
4. if the values at `left`, `mid`, and `right` are equal (duplicates), increment `left` and decrement `right`.
5. repeat the process until the pointers meet.
code example
here's how you could implement this in python:
explanation of the code
1. **initialization**: we set pointers `left` and `right` to the start and end of the list.
2. **while loop**: the loop continues as long as `l ...
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