What is the time complexity of Check if Array Is Sorted and Rotated?

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

What data structures are used to solve Check if Array Is Sorted and Rotated?

This problem uses Array. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Check if Array Is Sorted and Rotated?

Insight 1: A sorted array, when rotated, will have at most one point of discontinuity where the order changes from decreasing to increasing. This is because a single rotation only introduces one such change. Insight 2: We can efficiently detect this discontinuity by iterating through the array and counting the number of times a number is greater than the next number. If this count exceeds 1, it's not a rotated sorted array.

Which companies ask LeetCode 1752. Check if Array Is Sorted and Rotated in interviews?

This problem is frequently asked at SoundHound, tcs.

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Check if Array Is Sorted and Rotated - LeetCode 1752 Solution

Check if Array Is Sorted and Rotated - Complete Solution Guide

Check if Array Is Sorted and Rotated is LeetCode problem 1752, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an array nums , return true if the array was originally sorted in non-decreasing order, then rotated some number of positions (including zero) . Otherwise, return false . There may be duplicates in the original array. Note: An array A rotated by x positions results in an array B of the same length such that B[i] == A[(i+x) % A.length] for every valid index i . Example 1: Input: nums = [3,4,5,1,2] Output: true Explanation: [1,2,3,4,5] is the original sorted array. You can rotate the array b

Detailed Explanation

The problem asks whether a given array can be obtained by rotating a sorted array (in non-decreasing order). The input is an array of integers `nums`. The output is a boolean value: `true` if the array is a rotated version of a sorted array, and `false` otherwise. Duplicate numbers are allowed in the input array. The rotation can be zero positions (meaning the array is already sorted).

Solution Approach

The solution uses a single pass through the array to count the number of times an element is greater than its subsequent element. It leverages the modulo operator to wrap around from the end of the array to the beginning for the last element comparison. If this count is less than or equal to 1, it means the array could be a rotated version of a sorted array; otherwise, it's not.

Step-by-Step Algorithm

Step 1: Initialize a `count` variable to 0. This variable will track the number of times an element is greater than its successor.
Step 2: Iterate through the array using a `for` loop. For each element `nums[i]`, compare it to `nums[(i + 1) % n]`, where `n` is the length of the array. The modulo operator ensures correct comparison for the last element.
Step 3: If `nums[i] > nums[(i + 1) % n]`, increment the `count`.
Step 4: After the loop, check if `count <= 1`. If true, return `true` (rotated sorted array); otherwise, return `false`.

Key Insights

Insight 1: A sorted array, when rotated, will have at most one point of discontinuity where the order changes from decreasing to increasing. This is because a single rotation only introduces one such change.
Insight 2: We can efficiently detect this discontinuity by iterating through the array and counting the number of times a number is greater than the next number. If this count exceeds 1, it's not a rotated sorted array.
Insight 3: Handling the circularity of the array when comparing the last element with the first is crucial. The modulo operator (`%`) is used to elegantly handle this.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: SoundHound, tcs.

Check if Array Is Sorted and Rotated | LeetCode 1752

Easy

Array

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

Solution Code

Available in:

class Solution {
    public boolean check(int[] nums) {
        int n = nums.length;
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] > nums[(i + 1) % n]) {
                count++;
            }
        }
        return count <= 1;
    }
}

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Problem Info

Description

Problem 1752: Check if Array Is Sorted and Rotated - Detailed explanation and solution approaches.

Topics

Array

Companies

SoundHoundtcs

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:1752

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

4 steps

1
Step 1: Initialize a `count` variable to 0. This variable will track the number of times an element is greater than its successor.
2
Step 2: Iterate through the array using a `for` loop. For each element `nums[i]`, compare it to `nums[(i + 1) % n]`, where `n` is the length of the array. The modulo operator ensures correct comparison for the last element.
3
Step 3: If `nums[i] > nums[(i + 1) % n]`, increment the `count`.
4
Step 4: After the loop, check if `count <= 1`. If true, return `true` (rotated sorted array); otherwise, 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 algorithm iterates through the array once, performing a constant number of operations in each iteration. Therefore, the time complexity is linear. The algorithm uses only a constant amount of extra space to store the `count` variable and loop index. The space complexity is constant.

📖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

Sorting and Checking

Sort a copy of the input array. Then, check if the input array is a rotation of the sorted array. This involves iterating through all possible rotations.

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

Trade-offs: This approach is less efficient than the provided solution because of the sorting step. It's simpler to understand but less optimal in terms of time and space complexity. Use this only if you need a more straightforward, though less efficient, approach.

✨

Key Insights

1Insight 1: A sorted array, when rotated, will have at most one point of discontinuity where the order changes from decreasing to increasing. This is because a single rotation only introduces one such change.
2Insight 2: We can efficiently detect this discontinuity by iterating through the array and counting the number of times a number is greater than the next number. If this count exceeds 1, it's not a rotated sorted array.
3Insight 3: Handling the circularity of the array when comparing the last element with the first is crucial. The modulo operator (`%`) is used to elegantly handle this.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: An already sorted array (no rotation). The solution correctly identifies this as a rotated sorted array (count will be 0).
Edge case 2: An array that is a single rotation of a sorted array. The solution will find one discontinuity and return true.
Edge case 3: An array with all the same elements. The solution correctly identifies this as a rotated sorted array (count will be 0).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the circularity of the array when comparing the last element to the first. This can lead to incorrect results.

✗Common mistake 2: Assuming that only one peak implies a rotated sorted array; some other patterns might also have only one peak but are not rotated sorted arrays (e.g., a decreasing sequence). The given algorithm is robust in addressing this.

✅

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: Detecting if a circular buffer (like a rotating log file) is properly ordered after a rotation.

🏦Real-world scenario 2: Verifying the integrity of data that might have been circularly shifted during transmission or storage.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: October 11, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

❓

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 should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Check if Array Is Sorted and Rotated"?

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

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