LeetCode Top Interview 150

153. Find Minimum in Rotated Sorted Array

Medium

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

• [4,5,6,7,0,1,2] if it was rotated 4 times.
• [0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:

Input: nums = [3,4,5,1,2]

Output: 1

Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]

Output: 0

Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]

Output: 11

Explanation: The original array was [11,13,15,17] and it was rotated 4 times.

Constraints:

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• All the integers of nums are unique.
• nums is sorted and rotated between 1 and n times.

To solve the problem, we can use a binary search approach. This method leverages the properties of the rotated array and achieves the required O(log n) time complexity.

Steps:

1. Initialization:
   • Define two pointers, left and right, initialized to the start and end of the array, respectively.
2. Binary Search:
   • While left is less than right, perform the following steps:
     • Calculate the mid-point mid using integer division.
     • Determine if the middle element is greater than the rightmost element:
       • If nums[mid] is greater than nums[right], it means the minimum element is in the right half of the array. Move the left pointer to mid + 1.
       • Otherwise, the minimum element is in the left half or could be the mid itself. Move the right pointer to mid.
3. Return Result:
   • After the loop terminates, the left pointer will point to the minimum element in the array.

Implementation:

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1
        
        while left < right:
            mid = (left + right) // 2
            if nums[mid] > nums[right]:
                left = mid + 1
            else:
                right = mid
        
        return nums[left]

Explanation:

1. Initialization:
   • left starts at 0 and right starts at the last index of the array.
2. Binary Search:
   • The while loop runs as long as left is less than right.
   • mid is calculated as the average of left and right.
   • If the middle element nums[mid] is greater than nums[right], it means the minimum value is in the right part of the array. We discard the left half by setting left to mid + 1.
   • If nums[mid] is less than or equal to nums[right], it means the minimum value is in the left part, including mid. We set right to mid.
3. Return Result:
   • When the loop terminates, left will be at the index of the minimum element, so we return nums[left].

This approach efficiently narrows down the search space by half in each iteration, ensuring a logarithmic time complexity.

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