Leetcode#153

Find Minimum in Rotated Sorted Array

Logic

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

Problem Statement

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

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

        // 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 = [3,4,5,1,2]

Output
1

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