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Asked: 2024-09-24T22:09:04+05:30

You are given an array that was originally sorted in ascending order, but has been rotated at an unknown pivot. Your task is to find a specific target value within this rotated sorted array. If the target exists, return its index; if it does not, return -1. Consider how you can apply an efficient search algorithm to solve this problem, keeping in mind the properties of the rotated array. How would you approach this challenge and what is the expected time complexity of your solution?

anonymous user

I’ve been grappling with this interesting coding challenge and thought it’d be fun to see how others approach it! So, imagine you have an array that was once sorted in ascending order, but then it got rotated at some unknown pivot. Here’s the kicker: I’m trying to find a target value in this rotated array. If the value exists, I need to return its index; if not, I’ll just return -1.

Here’s a practical scenario: let’s say we have an array like this, maybe `[4, 5, 6, 7, 0, 1, 2]`. It’s clear that the algorithm to search through this isn’t just a simple linear search as it won’t be efficient, especially for larger arrays. We need to consider the properties of this rotated sorted array.

If we think about it, even though the array has been rotated, it still has a kind of order. There are two parts: the left side that’s still sorted up to the pivot and the right side that starts from the pivot and continues to the end. This means that during our search, we can use a modified binary search.

I’m wondering how you would tackle this problem. Probably involving some conditions while checking the mid-point of the array to determine which part is sorted – right? I can see the time complexity being around O(log n), which is much better than O(n).

I’m curious about the specifics of your approach. How would you implement this? Any code snippets or algorithms that come to your mind? And what edge cases would you consider—like when the array is empty or when it contains all duplicates? I’d love to hear your thoughts!

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    2 Answers

    1. anonymous user

      Okay, so this is a really cool problem! I’ve thought about it a bit and here’s how I would go about it.

      The idea is to use a modified binary search, like you mentioned! Here’s a basic outline of the approach:

      1. Start with two pointers, left and right, set to the beginning and end of the array.
      2. While left is less than or equal to right, calculate the mid index.
      3. Now, we check if the middle element is the target. If it is, return the mid index!
      4. If the left side is sorted (i.e., nums[left] <= nums[mid]), check if the target is within that range. If it is, narrow down the search to the left side; otherwise, search the right side.
      5. If the right side is sorted (i.e., nums[mid] <= nums[right]), do the opposite: check if the target is on that side and adjust accordingly.
      6. Repeat until you find the target or the search space is exhausted!

      Here’s a code snippet to illustrate this:

      
def search(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        
        # Check which part is sorted
        if nums[left] <= nums[mid]:  # Left side is sorted
            if nums[left] <= target < nums[mid]:
                right = mid - 1  # Target in left side
            else:
                left = mid + 1   # Target in right side
        else:  # Right side is sorted
            if nums[mid] < target <= nums[right]:
                left = mid + 1   # Target in right side
            else:
                right = mid - 1  # Target in left side
    return -1  # Not found

      As for edge cases, yeah! I’d check for:

      • An empty array: just return -1.
      • Arrays with all duplicates: the search should still work, but it may need special handling depending on if you’re able to tell apart duplicates from the target.
      • What if the array is already sorted (like [1, 2, 3])? The binary search still applies!

      This sounds like a fun challenge! Can’t wait to see how others tackle it too!

    2. anonymous user


      To tackle the problem of finding a target value in a rotated sorted array, we can leverage the properties of both the sorted and the rotated sections using a modified binary search approach. The key is to identify which half of the array is sorted at each step. We can start by initializing two pointers, `left` and `right`, to represent the bounds of our search. In each iteration, we calculate the midpoint `mid`. We’ll check whether the target value is equal to the element at `mid`. If it is, we return `mid`. If not, we need to determine which side of the array is sorted. If the left half, from `left` to `mid`, is sorted, we can check if the target lies within this range. If it does, we adjust our right pointer to `mid – 1`; otherwise, we move our left pointer to `mid + 1`. Conversely, if the right half is sorted, we apply the same logic to adjust the pointers accordingly.

      In terms of edge cases, we must consider scenarios such as an empty array, which should immediately return -1, or an array consisting of all duplicate values where the target might be present or absent. These cases should be handled gracefully by our algorithm to ensure robustness. The time complexity of this approach is O(log n), making it significantly more efficient than a linear search. Here’s a simple implementation in Python to illustrate the approach:

      def search(nums, target):
    left, right = 0, len(nums) - 1
    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        if nums[left] <= nums[mid]:  # left half is sorted
            if nums[left] <= target < nums[mid]:
                right = mid - 1
            else:
                left = mid + 1
        else:  # right half is sorted
            if nums[mid] < target <= nums[right]:
                left = mid + 1
            else:
                right = mid - 1
    return -1


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