LeetCode Link: 153. Find Minimum in Rotated Sorted Array
Language: C#
Problem Statement
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.
Examples
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.
Solution
public class Solution
{
public int FindMin(int[] nums)
{
var left = 0;
var right = nums.Length - 1;
if (nums[0] < nums[right])
{
return nums[0];
}
while (left <= right)
{
var mid = (left + right) / 2;
if (mid > 0 && nums[mid] < nums[mid - 1] )
{
return nums[mid];
}
if (mid < nums.Length - 1 && nums[mid] > nums[mid + 1])
{
return nums[mid + 1];
}
if (nums[left] > nums[mid])
{
right = mid - 1;
}
else
{
left = mid + 1;
}
}
return nums[0];
}
}
Complexity
Time Complexity: O(logN)
Space Complexity: O(1)