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

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

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

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

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

Insight 1: The presence of duplicates makes it impossible to always determine if the left or right half is sorted in O(1) time. When `nums[left] == nums[mid]`, we can't know which side is sorted, leading to a worst-case linear time complexity. Insight 2: We can still use a modified binary search. If `nums[left] nums[mid]`, the right half is sorted. If `nums[left] == nums[mid]`, we increment `left` to shrink the search space.

Which companies ask LeetCode 81. Search in Rotated Sorted Array II in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Meta and 3 more companies.

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Search in Rotated Sorted Array II - LeetCode 81 Solution

Q: How to solve Search in Rotated Sorted Array II?

The provided code implements a modified binary search algorithm to search for the target in the rotated sorted array. It maintains two pointers, `left` and `right`, initially pointing to the start and end of the array. In each iteration, it calculates the middle index `mid`. It checks if `nums[mid]` is equal to the target. If not, it determines which half of the array is sorted and checks if the target lies within that sorted half. If it does, it updates the `right` pointer; otherwise, it updates the `left` pointer. A crucial step is handling the case where `nums[left] == nums[mid]`. In this situation, we increment `left` by 1 to shrink the search window, effectively skipping the duplicate and potentially revealing the true sorted half in subsequent iterations.

Search in Rotated Sorted Array II - Complete Solution Guide

Search in Rotated Sorted Array II is LeetCode problem 81, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values). Before being passed to your function, nums is rotated at an unknown pivot index k ( 0 <= k < nums.length ) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] ( 0-indexed ). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4] . Given the array nums after the rotation and an integer target , retu

Detailed Explanation

The problem asks us to search for a target value in a rotated sorted array. The array was initially sorted in non-decreasing order but has been rotated at an unknown pivot index. The key challenge is that the array may contain duplicate values, which complicates the standard binary search approach. We need to return `true` if the target is found in the array, and `false` otherwise.

Solution Approach

The provided code implements a modified binary search algorithm to search for the target in the rotated sorted array. It maintains two pointers, `left` and `right`, initially pointing to the start and end of the array. In each iteration, it calculates the middle index `mid`. It checks if `nums[mid]` is equal to the target. If not, it determines which half of the array is sorted and checks if the target lies within that sorted half. If it does, it updates the `right` pointer; otherwise, it updates the `left` pointer. A crucial step is handling the case where `nums[left] == nums[mid]`. In this situation, we increment `left` by 1 to shrink the search window, effectively skipping the duplicate and potentially revealing the true sorted half in subsequent iterations.

Step-by-Step Algorithm

Step 1: Initialize `left` to 0 and `right` to `nums.length - 1`.
Step 2: While `left` is less than or equal to `right`, repeat steps 3-8.
Step 3: Calculate the middle index `mid` as `left + (right - left) // 2`.
Step 4: If `nums[mid]` equals the target, return `true`.
Step 5: If `nums[left]` equals `nums[mid]`, increment `left` by 1 and continue to the next iteration. This handles the duplicate case where we can't determine which side is sorted immediately.
Step 6: If `nums[left]` is less than `nums[mid]`, the left part of the array is sorted. Check if the target lies within this sorted portion (i.e., `nums[left] <= target < nums[mid]`). If it does, update `right` to `mid - 1`; otherwise, update `left` to `mid + 1`.
Step 7: Otherwise (if `nums[left]` is greater than `nums[mid]`), the right part of the array is sorted. Check if the target lies within this sorted portion (i.e., `nums[mid] < target <= nums[right]`). If it does, update `left` to `mid + 1`; otherwise, update `right` to `mid - 1`.
Step 8: If the loop completes without finding the target, return `false`.

Key Insights

Insight 1: The presence of duplicates makes it impossible to always determine if the left or right half is sorted in O(1) time. When `nums[left] == nums[mid]`, we can't know which side is sorted, leading to a worst-case linear time complexity.
Insight 2: We can still use a modified binary search. If `nums[left] < nums[mid]`, the left half is sorted. If `nums[left] > nums[mid]`, the right half is sorted. If `nums[left] == nums[mid]`, we increment `left` to shrink the search space.
Insight 3: The core of the solution lies in deciding whether to search in the left or right half based on whether the target lies within the sorted portion of the array.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Meta, Microsoft, Uber, Yahoo.

Search in Rotated Sorted Array II | LeetCode 81

Medium

Array Binary Search

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

Solution Code

Available in:

class Solution {
    public boolean search(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;

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

            if (nums[mid] == target) {
                return true;
            }

            if (nums[left] == nums[mid]) {
                left++;
                continue;
            }

            // Left part is sorted
            if (nums[left] < nums[mid]) {
                if (nums[left] <= target && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
            // Right part is sorted
            else {
                if (nums[mid] < target && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 81: Search in Rotated Sorted Array II - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

AdobeAmazonAppleBloombergMetaMicrosoftUberYahoo

Stats

Difficulty:Medium

Languages:4

Companies:8

Problem #:81

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

8 steps

1
Step 1: Initialize `left` to 0 and `right` to `nums.length - 1`.
2
Step 2: While `left` is less than or equal to `right`, repeat steps 3-8.
3
Step 3: Calculate the middle index `mid` as `left + (right - left) // 2`.
4
Step 4: If `nums[mid]` equals the target, return `true`.
5
Step 5: If `nums[left]` equals `nums[mid]`, increment `left` by 1 and continue to the next iteration. This handles the duplicate case where we can't determine which side is sorted immediately.
6
Step 6: If `nums[left]` is less than `nums[mid]`, the left part of the array is sorted. Check if the target lies within this sorted portion (i.e., `nums[left] <= target < nums[mid]`). If it does, update `right` to `mid - 1`; otherwise, update `left` to `mid + 1`.
7
Step 7: Otherwise (if `nums[left]` is greater than `nums[mid]`), the right part of the array is sorted. Check if the target lies within this sorted portion (i.e., `nums[mid] < target <= nums[right]`). If it does, update `left` to `mid + 1`; otherwise, update `right` to `mid - 1`.
8
Step 8: If the loop completes without finding the target, return `false`.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is primarily determined by the number of iterations in the `while` loop. In the ideal case (no duplicates), the binary search halves the search space in each step, leading to logarithmic time complexity. However, the presence of duplicates can degrade this to linear time complexity. The space complexity is constant because the algorithm's memory usage doesn't grow 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 compare each element with the target.

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

Trade-offs: Simpler to implement, but less efficient than binary search, especially for large arrays, even when the binary search degrades to O(n).

✨

Key Insights

1Insight 1: The presence of duplicates makes it impossible to always determine if the left or right half is sorted in O(1) time. When `nums[left] == nums[mid]`, we can't know which side is sorted, leading to a worst-case linear time complexity.
2Insight 2: We can still use a modified binary search. If `nums[left] < nums[mid]`, the left half is sorted. If `nums[left] > nums[mid]`, the right half is sorted. If `nums[left] == nums[mid]`, we increment `left` to shrink the search space.
3Insight 3: The core of the solution lies in deciding whether to search in the left or right half based on whether the target lies within the sorted portion of the array.

📚

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 (`nums.length == 0`). The provided code implicitly handles this case because the `while` loop condition `left <= right` will be false from the start, and the function will return `false`.
Edge case 2: Target is at the beginning or end of the array. The code correctly handles this because the binary search will eventually converge to the index of the target (if it exists) and return `true`.
Edge case 3: Array is not rotated (already sorted). The code still works correctly as the binary search logic is valid even for sorted arrays.
Edge case 4: All elements are the same and the target is present. The `nums[left] == nums[mid]` condition will be repeatedly met, incrementing left until `nums[mid]` equals the target. However, if the target is not present, the loop might take longer to complete.
Edge case 5: All elements are the same and target is not present. The `nums[left] == nums[mid]` causes left to increment and the algorithm eventually terminates returning false.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the `nums[left] == nums[mid]` case correctly, leading to an infinite loop or incorrect results.

✗Common mistake 2: Incorrectly determining which half of the array is sorted due to the rotation, resulting in searching the wrong portion of the array.

✗Common mistake 3: Failing to account for the possibility of duplicates when designing the binary search logic. Many standard binary search implementations assume unique elements.

✅

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: Searching for data in a database table that is partially sorted but has undergone some transformations or updates that disrupted the sorting order.

🏦Real-world scenario 2: Implementing a search functionality in a log file where the entries are mostly sorted by timestamp, but occasional out-of-order entries can occur.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: August 21, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

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

For this medium-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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