Find Minimum in Rotated Sorted Array II
LeetCode 154 | Difficulty: Hardβ
HardProblem Descriptionβ
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?
Topics: Array, Binary Search
Approachβ
Binary Searchβ
Binary search reduces the search space by half at each step. The key insight is identifying the monotonic property β what condition lets you decide to go left or right?
Sorted array, or searching for a value in a monotonic function/space.
Solutionsβ
Solution 1: C# (Best: 112 ms)β
| Metric | Value |
|---|---|
| Runtime | 112 ms |
| Memory | N/A |
| Date | 2018-07-04 |
public class Solution {
public int FindMin(int[] nums) {
int lo = 0, hi = nums.Length - 1;
while (lo < hi)
{
int mid = lo + (hi - lo) / 2;
if (nums[mid] > nums[hi])
{
lo = mid+1;
}
else if (nums[mid] < nums[lo]){
hi = mid;
}
else
{
hi--;
}
}
return nums[lo];
}
}
Complexity Analysisβ
| Approach | Time | Space |
|---|---|---|
| Binary Search | $O(log n)$ | $O(1)$ |
Interview Tipsβ
- Break the problem into smaller subproblems. Communicate your approach before coding.
- Precisely define what the left and right boundaries represent, and the loop invariant.