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, return the minimum element of this array.
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:
Approach1 O(n):
class Solution {
public int findMin(int[] nums) {
if(nums.length==1) return nums[0];
for(int i=0;i<nums.length;i++)
{
if(nums[i]>nums[(i+1)%nums.length])
{
return nums[(i+1)%nums.length];
}
}
return -1;
}
}Approach 2 O(log n) Binary Search:
class Solution {
public int findMin(int[] nums) {
if(nums.length==1) return nums[0];
int start = 0, end=nums.length-1;
if(nums[0]<nums[end]) return nums[0];
while(start<=end)
{
int mid = start+(end-start)/2;
if(nums[mid+1]<nums[mid]) return nums[mid+1];
if(nums[mid-1]>nums[mid]) return nums[mid];
if(nums[mid]>nums[0])
{
start = mid+1;
}
else end = mid-1;
}
return -1;
}
}