Find Minimum in Rotated Sorted Array II 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 rotated4
times.[0,1,4,4,5,6,7]
if it was rotated7
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 between1
andn
times.
Find Minimum in Rotated Sorted Array II Solutions
✅Time: O(logn)→O(n)
✅Space: O(1)
C++
class Solution {
public:
int findMin(vector<int>& nums) {
int l = 0;
int r = nums.size() - 1;
while (l < r) {
const int m = l + (r - l) / 2;
if (nums[m] == nums[r])
--r;
else if (nums[m] < nums[r])
r = m;
else
l = m + 1;
}
return nums[l];
}
};
Java
class Solution {
public int findMin(int[] nums) {
int l = 0;
int r = nums.length - 1;
while (l < r) {
final int m = l + (r - l) / 2;
if (nums[m] == nums[r])
--r;
else if (nums[m] < nums[r])
r = m;
else
l = m + 1;
}
return nums[l];
}
}
Python
class Solution:
def findMin(self, nums: List[int]) -> int:
l = 0
r = len(nums) - 1
while l < r:
m = (l + r) // 2
if nums[m] == nums[r]:
r -= 1
elif nums[m] < nums[r]:
r = m
else:
l = m + 1
return nums[l]
Watch Tutorial
Checkout more Solutions here