Leetcode#154

Find Minimum in Rotated Sorted Array II With Duplicates

Problem Statement

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 1:

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

Example 2:

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

Constraints:

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• nums is sorted and rotated between 1 and n times.

Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

Logic

• Find min in rotated array logic is same for this solution
• Plus we added extra logic to shrink the array.
• if we find left , mid and right value are equal,
  • Just shrink the array by 1
  • left++ & right — ; continue
• Please read normal Binary search code for finding target and its comparison with rotated array.
• This will help to build foundation.
  • Binary Search vs Rotated Binary Search

Golang Solution

func findMin(nums []int) int {

	n := len(nums)
	left, right := 0, n-1

	// Check which part of the array to search
	for left < right {
		middle := left + (right-left)/2

		// Handle duplicates by shrinking the bounds
        if nums[left] == nums[middle] && nums[middle] == nums[right] {
            left++
            right--
            continue
        }

		// Minimum must be in the right half
		if nums[middle] > nums[right] {
			left = middle + 1
		} else {
			// Minimum is in the left half or at middle
			right = middle
		}
	}

	// The left pointer will point to the minimum
	return nums[left]
}

Output

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

Output
0

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