153. Find Minimum in Rotated Sorted Array

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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.

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.

Solutions

Solution 1: Binary Search

We can use binary search to solve this problem.

First, we define two pointers $l$ and $r$ pointing to the start and end of the array respectively. Then we enter a loop until $l$ is no longer less than $r$.

In each iteration, we calculate the middle position $mid$ and compare $nums[mid]$ with $nums[n-1]$. If $nums[mid]$ is greater than $nums[n-1]$, the minimum value is to the right of $mid$, so we update $l$ to $mid + 1$. Otherwise, the minimum value is at $mid$ or to its left, so we update $r$ to $mid$. When the loop ends, pointer $l$ will point to the minimum value, and we return $nums[l]$.

The time complexity is $O(\log n)$, where $n$ is the length of array $\textit{nums}$. The space complexity is $O(1)$.

Python3

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

Java

class Solution {
    public int findMin(int[] nums) {
        int l = 0, r = nums.length - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] > nums[nums.length - 1]) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        return nums[l];
    }
}

C++

class Solution {
public:
    int findMin(vector<int>& nums) {
        int l = 0, r = nums.size() - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] > nums.back()) {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        return nums[l];
    }
};

Go

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

TypeScript

function findMin(nums: number[]): number {
    let l = 0,
        r = nums.length - 1;
    while (l < r) {
        let mid = (l + r) >> 1;
        if (nums[mid] > nums[nums.length - 1]) {
            l = mid + 1;
        } else {
            r = mid;
        }
    }
    return nums[l];
}

Rust

impl Solution {
    pub fn find_min(nums: Vec<i32>) -> i32 {
        let (mut l, mut r) = (0, nums.len() - 1);
        while l < r {
            let mid = (l + r) >> 1;
            if nums[mid] > nums[nums.len() - 1] {
                l = mid + 1;
            } else {
                r = mid;
            }
        }
        nums[l]
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function (nums) {
    let l = 0,
        r = nums.length - 1;
    while (l < r) {
        let mid = (l + r) >> 1;
        if (nums[mid] > nums[nums.length - 1]) {
            l = mid + 1;
        } else {
            r = mid;
        }
    }
    return nums[l];
};