153. Find Minimum in Rotated Sorted Array

LeetCode 153. Find Minimum in Rotated Sorted Array

Description

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.

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.

Tags

Array, Binary Search

Solution

Perform binary search to find the precipice in the array, by comparing nums[mid] with the last element of the array. If nums[mid] is larger then move the left pointer to the middle index, otherwise move the right pointer. Finally, we return the smaller one of both nums[start] and nums[end].

Complexity

• Time complexity: $O(\log(n))$

• Space complexity: $O(1)$

Code

func findMin(nums []int) int {
	start, end := 0, len(nums)-1
	for start+1 < end {
		mid := start + (end-start)/2
		if nums[mid] >= nums[end] {
			start = mid
		} else {
			end = mid
		}
	}
	if nums[start] < nums[end] {
		return nums[start]
	}
	return nums[end]
}