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

The optimal solution has a time complexity of O(log n).

What data structures are used to solve Find Minimum in Rotated Sorted Array II?

This problem uses Array, Binary Search. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find Minimum in Rotated Sorted Array II?

Insight 1: The minimum element will always be the one that is smaller than its preceding element. If the array is not rotated, the first element is the minimum. Insight 2: Binary search can be adapted to efficiently search through the rotated array. The standard binary search needs to be modified to handle the rotated and duplicate array conditions.

Which companies ask LeetCode 154. Find Minimum in Rotated Sorted Array II in interviews?

This problem is frequently asked at Google.

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Find Minimum in Rotated Sorted Array II - LeetCode 154 Solution

Find Minimum in Rotated Sorted Array II - Complete Solution Guide

Find Minimum in Rotated Sorted Array II is LeetCode problem 154, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 ov

Detailed Explanation

The problem asks us to find the minimum element in a rotated sorted array that may contain duplicate elements. A rotated sorted array is an array that was initially sorted in ascending order but has been shifted a certain number of positions to the right. For example, `[0, 1, 2, 4, 5, 6, 7]` might become `[4, 5, 6, 7, 0, 1, 2]`. The key challenge here is the presence of duplicates, which can make it difficult to determine which half of the array contains the minimum element using a simple binary search.

Solution Approach

The solution uses a modified binary search approach. We maintain two pointers, `left` and `right`, initially pointing to the start and end of the array, respectively. In each iteration, we calculate the middle index `mid`. We compare `nums[mid]` with `nums[right]` to determine which half potentially contains the minimum element. If `nums[mid] > nums[right]`, it means the minimum element is in the right half (from `mid + 1` to `right`). If `nums[mid] < nums[right]`, the minimum element is in the left half (from `left` to `mid`, inclusive). If `nums[mid] == nums[right]`, we decrement `right` by 1, effectively discarding the rightmost element. This step is crucial for handling duplicates and ensuring the algorithm converges.

Step-by-Step Algorithm

Step 1: Initialize `left` to 0 and `right` to `nums.length - 1`.
Step 2: While `left < right`:
Step 3: Calculate `mid` as `left + (right - left) // 2` to prevent potential integer overflow.
Step 4: If `nums[mid] > nums[right]`, then `left = mid + 1` (minimum is in the right half).
Step 5: Else if `nums[mid] < nums[right]`, then `right = mid` (minimum is in the left half, including mid).
Step 6: Else (if `nums[mid] == nums[right]`), then `right--` (discard the rightmost element).
Step 7: After the loop finishes, `left` and `right` will be equal, pointing to the minimum element. Return `nums[left]`.

Key Insights

Insight 1: The minimum element will always be the one that is smaller than its preceding element. If the array is not rotated, the first element is the minimum.
Insight 2: Binary search can be adapted to efficiently search through the rotated array. The standard binary search needs to be modified to handle the rotated and duplicate array conditions.
Insight 3: Duplicates can hinder the effectiveness of binary search. If `nums[mid] == nums[right]`, it becomes impossible to determine which half contains the minimum using a simple comparison. In this case, we can safely discard `nums[right]` because if `nums[right]` were the minimum, then `nums[mid]` would have to be strictly less than it, which contradicts `nums[mid] == nums[right]`.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Google.

Find Minimum in Rotated Sorted Array II | LeetCode 154

Hard

Array Binary Search

Time: O(log n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int findMin(int[] nums) {
        int left = 0;
        int right = nums.length - 1;

        while (left < right) {
            int mid = left + (right - left) / 2;

            if (nums[mid] > nums[right]) {
                // The minimum element is in the right half
                left = mid + 1;
            } else if (nums[mid] < nums[right]) {
                // The minimum element is in the left half, including mid
                right = mid;
            } else { // nums[mid] == nums[right]
                // We can't determine which half is sorted.
                // However, we can safely discard the rightmost element.
                // This is because if nums[right] were the minimum,
                // nums[mid] would be greater than it (in a rotated array),
                // unless nums[mid] is also the minimum.
                // In the case they are equal, we can discard one of them
                // without losing the minimum value.
                right--;
            }
        }

        return nums[left];
    }
}

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Problem Info

Description

Problem 154: Find Minimum in Rotated Sorted Array II - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

Google

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:154

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `left` to 0 and `right` to `nums.length - 1`.
2
Step 2: While `left < right`:
3
Step 3: Calculate `mid` as `left + (right - left) // 2` to prevent potential integer overflow.
4
Step 4: If `nums[mid] > nums[right]`, then `left = mid + 1` (minimum is in the right half).
5
Step 5: Else if `nums[mid] < nums[right]`, then `right = mid` (minimum is in the left half, including mid).
6
Step 6: Else (if `nums[mid] == nums[right]`), then `right--` (discard the rightmost element).
7
Step 7: After the loop finishes, `left` and `right` will be equal, pointing to the minimum element. Return `nums[left]`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log n)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is logarithmic when the binary search effectively halves the search space in each iteration. However, with many duplicates, the `right--` operation degrades the performance, leading to a linear time complexity in the worst case. The space complexity is constant because the algorithm operates in place and does not use any auxiliary data structures that scale with the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Linear Search

Iterate through the array and keep track of the minimum element encountered so far.

Time: O(n)Space: O(1)

Trade-offs: Simpler to implement but less efficient than binary search when duplicates are not prevalent. Useful if the overhead of binary search is not warranted or if the array is expected to be small.

✨

Key Insights

1Insight 1: The minimum element will always be the one that is smaller than its preceding element. If the array is not rotated, the first element is the minimum.
2Insight 2: Binary search can be adapted to efficiently search through the rotated array. The standard binary search needs to be modified to handle the rotated and duplicate array conditions.
3Insight 3: Duplicates can hinder the effectiveness of binary search. If `nums[mid] == nums[right]`, it becomes impossible to determine which half contains the minimum using a simple comparison. In this case, we can safely discard `nums[right]` because if `nums[right]` were the minimum, then `nums[mid]` would have to be strictly less than it, which contradicts `nums[mid] == nums[right]`.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Empty array. While not explicitly mentioned in the constraints, it's good practice to consider it. The given code would likely throw an exception if the input array is empty because of accessing `nums[right]`. A check `if (nums.length == 0) return -1;` could be added at the beginning for robustness (returning -1 or throwing an exception depending on the requirements).
Edge case 2: Array with only one element. The algorithm works correctly in this case, as the `while` loop condition `left < right` will be false initially, and the function will directly return `nums[0]`.
Edge case 3: Array with all elements the same. The algorithm handles this case correctly due to the `right--` step in the `else` condition. This ensures the loop eventually terminates and returns the correct (and only) value.
Edge case 4: Array is already sorted and not rotated (e.g., [1, 2, 3, 4, 5]). The algorithm correctly identifies the first element as the minimum.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where `nums[mid] == nums[right]`. Without the `right--` step, the algorithm can get stuck in an infinite loop if there are many duplicates.

✗Common mistake 2: Incorrectly updating `left` or `right` in the binary search. For instance, setting `right = mid - 1` when the minimum might be at `mid`, or setting `left = mid` when the minimum is known to be to the right.

✗Common mistake 3: Integer overflow when calculating `mid`. Using `mid = (left + right) / 2` can lead to overflow if `left + right` is a large number. The recommended approach is `mid = left + (right - left) // 2`.

✗Common mistake 4: Assuming all elements are distinct, and not accounting for the edge cases that duplicates introduce.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Database systems often use sorted data structures to efficiently search for records. Rotated sorted arrays can model situations where data is partially updated or reordered, and finding the minimum is equivalent to finding the starting point of the sorted sequence.

🏦Real-world scenario 2: Load balancing in distributed systems. You might need to find the server with the minimum load after a series of rotations or updates to the load distribution.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find Minimum in Rotated Sorted Array II"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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