# 154. Find Minimum in Rotated Sorted Array II

## Description

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

## Solutions

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

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

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


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

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


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