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.

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

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

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

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


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

• function findMin(nums: number[]): number {
let left = 0;
let right = nums.length - 1;
while (left < right) {
const mid = (left + right) >>> 1;
if (nums[mid] > nums[right]) {
left = mid + 1;
} else {
right = mid;
}
}
return nums[left];
}


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


• impl Solution {
pub fn find_min(nums: Vec<i32>) -> i32 {
let mut left = 0;
let mut right = nums.len() - 1;
while left < right {
let mid = left + (right - left) / 2;
if nums[mid] > nums[right] {
left = mid + 1;
} else {
right = mid;
}
}
nums[left]
}
}