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

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Approach

Binary Search

Binary search reduces the search space by half at each step. The key insight is identifying the monotonic property — 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.

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Solutions

Solution 1: C# (Best: 112 ms)

Metric	Value
Runtime	112 ms
Memory	N/A
Date	2018-07-04

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];
    }
}

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Complexity Analysis

Approach	Time	Space
Binary Search	$O(log n)$	$O(1)$

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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.