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Find Minimum in Rotated Sorted Array II

LeetCode 154 | Difficulty: Hard​

Hard

Problem Description​

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


Approach​

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?

When to use

Sorted array, or searching for a value in a monotonic function/space.


Solutions​

Solution 1: C# (Best: 112 ms)​

MetricValue
Runtime112 ms
MemoryN/A
Date2018-07-04
Solution
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 Analysis​

ApproachTimeSpace
Binary Search$O(log n)$$O(1)$

Interview Tips​

Key Points
  • Break the problem into smaller subproblems. Communicate your approach before coding.
  • Precisely define what the left and right boundaries represent, and the loop invariant.