Problem Description

The problem presents an array that has been sorted in ascending order but then has been rotated between 1 and n times. Rotating the array means moving the last element to the front. For example, if we have an array [1, 2, 3, 4, 5] and rotate it once, it becomes [5, 1, 2, 3, 4]. The task is to find the minimum element in this rotated array efficiently, even though the array may contain duplicates.

Intuition

Given that the array is initially sorted before the rotations, we can use binary search to find the minimum element. Here's the thought process:

• If the middle element is greater than the rightmost element, the smallest value must be to the right of the middle. This is because the elements are originally sorted in ascending order, so if there's a large number in the middle, the array must have been rotated at some point after that number.
• If the middle element is less than the rightmost element, the smallest value must be at the middle or to the left of the middle.
• If the middle element equals the rightmost element, we can't determine the position of the minimum element, but we can safely reduce the search space by ignoring the rightmost element since even if it is the minimum, it will exist as a duplicate elsewhere in the list.

The solution consistently halves the search interval, which means that the time complexity would be better than linear - in fact, it's O(log n) in the best case. However, when duplicates are present, and we encounter the same values at both the middle and end of our search range, we must decrease our search range in smaller steps, which in the worst case could result in an O(n) time complexity when all elements are duplicates. The approach is both practical and efficient for most cases.

Learn more about Binary Search patterns.

Solution Approach

The solution uses a classic binary search pattern to find the minimum element. The key insight of binary search is to repeatedly divide the search interval in half. If the interval can be divided, it suggests that through a comparison in each step, one can eliminate half of the remaining possibilities. Here's how it is applied in this context:

• We start by setting two pointers, left at 0 and right at len(nums) - 1, to represent the search space's bounds.
• While left is less than right, we are not done searching:
  • We calculate mid by taking the average of left and right, effectively dividing the search space in half.
  • If nums[mid] > nums[right], we know that the smallest value must be to the right of mid, so we set left to mid + 1.
  • If nums[mid] < nums[right], we then know that the smallest value is either at mid or to its left. So, we move right to mid.
  • If nums[mid] is equal to nums[right], we cannot be certain where the smallest value is, but we can reduce the search space by decrements right by 1. This is because even if nums[right] is the smallest value, it will not be lost as the property of the array being sorted (aside from the rotation) means a duplicate must exist to the left.
• Once left equals right, the loop terminates and left (or right, since they are equal) points to the smallest element.

The code handles the presence of duplicates gracefully, ensuring the search space is narrowed even when the value of nums[mid] is equal to nums[right]. While binary search usually has an O(log n) complexity, this variant's complexity can degrade to O(n) in the worst-case scenario when there are many identical elements.

The reason why this approach is efficient is because it strives to minimize the number of elements considered at each step, allowing for an early exit in many cases as compared to a brute-force linear search that would always take O(n) time.

Example Walkthrough

Let's consider the rotated sorted array nums = [4, 5, 6, 7, 0, 1, 2] and walk through how the solution approach finds the minimum element.

1. Set the left pointer to 0 and the right pointer to 6, since len(nums) - 1 = 6. Our array now looks like this: 4, 5, 6, 7, 0, 1, 2, with left pointing to 4 and right pointing to 2.

2. Since left is less than right, we calculate the middle index mid = (left + right) // 2, which is (0 + 6) // 2 = 3. The value at this index is 7.

3. Now we compare the middle element nums[mid] with nums[right]:

   • nums[mid] is 7 and nums[right] is 2. Since 7 > 2, we know the smallest element is to the right of mid. We set left to mid + 1, which makes left 4.

4. The array between indices 4 and 6 is now our search space: 0, 1, 2. We repeat step 2 and find a new middle at mid = (4 + 6) // 2 = 5. The number at index 5 is 1.

5. We now compare the nums[mid] with nums[right] again:

   • nums[mid] is 1 and nums[right] is 2. Since 1 < 2, we know the smallest element is at mid or to the left of mid. We move right to mid, making right 5.

6. Our search space is now just [0, 1]. Calculating mid gives us (4 + 5) // 2 = 4. At nums[mid], we have a 0.

7. We compare nums[mid] with nums[right]:

   • nums[mid] is 0 and nums[right] is 1. Since 0 < 1, we continue to narrow the search and move right to mid, leaving us with right being 4.

8. With both left and right pointing to the same index, which is 4, the loop terminates. We have left equals right equals 4, and nums[left] or nums[right] gives us the minimum element, which is 0.

In this example, we successfully found the minimum element of the rotated sorted array using the binary search method, demonstrating the efficiency of the approach. The key takeaway is the decision-making at each step to reduce the search space, which is much quicker than a linear search especially when dealing with a larger array.

Solution Implementation

Time and Space Complexity

The provided code snippet is designed to find the minimum element in a rotated sorted array, handling duplicates. The algorithm employs a binary search technique.

Time Complexity:

The worst-case time complexity of this code is O(n). In the average and best case, where the majority of elements are not duplicates, it approaches O(log n). However, in the worst case, when the algorithm must decrement the right pointer one by one due to the presence of identical elements at the end of the array (else: right -= 1), the complexity degrades to O(n). This happens when duplicates are present and cannot be ruled out by a regular binary search.

Space Complexity:

The space complexity of the code is O(1). No additional space is utilized that is dependent on the input size; only a constant amount of extra space is used for variables like left, right, and mid.

Learn more about how to find time and space complexity quickly using problem constraints.