154. Find Minimum in Rotated Sorted Array II

Problem Statement

link: https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/ https://leetcode.cn/problems/find-minimum-in-rotated-sorted-array-ii/

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

[4,5,6,7,0,1,4] if it was rotated 4 times. [0,1,4,4,5,6,7] 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 that may contain duplicates, return the minimum element of this array.

You must decrease the overall operation steps as much as possible. Example:

Input: nums = [1,3,5] Output: 1

Input: nums = [2,2,2,0,1] Output: 0

Solution Approach

The solution efficiently identifies the minimum element using modified binary search tailored for rotated arrays.

Algorithm

1. Initialization: Set pointers l and r to the start and end of the array.
2. Midpoint Evaluation: Check the middle of the array; if it’s less than the rightmost value, the smallest value is in the left half; if it’s greater, the smallest value is in the right half.
3. Handling Duplicates: If the middle value equals the rightmost value, it’s not clear which half of the array the minimum is in. Decrement the right pointer and continue the search.

Implement

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

    
    

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