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Find Minimum in Rotated Sorted Array II

Hard Binary Search Rotated Array Original on LeetCode

Find Minimum in Rotated Sorted Array II is a hard binary search problem solved with the rotated array pattern. The best approach, optimal (binary search, shrink on ties), runs in O(log n) average, O(n) worst time and O(1) space. Below are 2 approaches in Java, from linear scan up.

Problem

A sorted array that may contain duplicates was rotated at an unknown pivot. Return its minimum, doing as little work as possible.

Examples

Example 1

Input
nums = [3, 3, 1, 3]
Output
1

Example 2

Input
nums = [2, 2, 2, 0, 1]
Output
0

Constraints

  • 1 <= nums.length <= 5000; duplicates allowed.

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
Linear scanO(n)O(1)
Optimal (binary search, shrink on ties)O(log n) average, O(n) worstO(1)

1Linear scan

TimeO(n)
SpaceO(1)

Return the smallest element. With duplicates, this is also the worst case of any method.

  1. Track the minimum.
Java
class Solution {
    public int findMin(int[] nums) {
        int min = nums[0];
        for (int x : nums) min = Math.min(min, x);
        return min;
    }
}

2Optimal (binary search, shrink on ties)

TimeO(log n) average, O(n) worstArrays like [1,1,1,1] force the hi-- step n times.
SpaceO(1)

Same as the distinct version, with one extra case. If nums[mid] == nums[hi], nums[hi] has a copy at mid, so dropping hi keeps the minimum in range: hi--.

  1. nums[mid] > nums[hi]: lo = mid + 1.
  2. nums[mid] < nums[hi]: hi = mid.
  3. Equal: hi--.
Java
class Solution {
    public int findMin(int[] nums) {
        int lo = 0, hi = nums.length - 1;
        while (lo < hi) {
            int mid = (lo + hi) >>> 1;
            if (nums[mid] > nums[hi]) lo = mid + 1;
            else if (nums[mid] < nums[hi]) hi = mid;
            else hi--;
        }
        return nums[lo];
    }
}

Edge cases to test

  • All values equal
  • nums[mid] == nums[hi] with the minimum on either side

Hints

Hint 1

When nums[mid] == nums[hi] you cannot tell which side the drop is on. But you can safely discard hi.

FAQ

What is the best time complexity for Find Minimum in Rotated Sorted Array II?

Optimal (binary search, shrink on ties) runs in O(log n) average, O(n) worst time and O(1) extra space. Arrays like [1,1,1,1] force the hi-- step n times.

Which pattern does Find Minimum in Rotated Sorted Array II use?

It is a binary search problem that uses the rotated array pattern. Other problems with the same pattern: Find How Many Times Array is Rotated, Find Minimum in Rotated Sorted Array, Search in Rotated Sorted Array.

Is there a brute force solution for Find Minimum in Rotated Sorted Array II?

Yes. Linear scan takes O(n) time and O(1) space. Return the smallest element.

Which edge cases should I test for Find Minimum in Rotated Sorted Array II?

All values equal; nums[mid] == nums[hi] with the minimum on either side.