LeetCode Problem
154. Find Minimum in Rotated Sorted Array II
Link to LeetCode
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 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
Analysis
We only need to add one more condition, which checks if the left-most element and the right-most element are equal. If they are we can simply drop one of them. In my solution below, I drop the left element whenever the left-most equals to the right-most.
class Solution {
public int findMin(int[] nums) {
int start = 0, end = nums.length-1;
while (start < end) {
int mid = start + (end - start) / 2;
if (nums[mid] > nums[end]) {
start = mid+1;
} else if (nums[mid] < nums[end]) {
end = mid;
} else {
end--;
}
}
return nums[end];
}
}