What is the time complexity of Rotate Array?

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

What data structures are used to solve Rotate Array?

This problem uses Array, Math, Two Pointers. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Rotate Array?

Insight 1: Using the modulo operator (%) is crucial to handle cases where 'k' is greater than the length of the array 'nums'. This ensures that the rotation happens within the bounds of the array. Insight 2: The problem can be efficiently solved in-place by reversing array segments. Reversing the entire array, then reversing the first k elements, and finally reversing the remaining elements will effectively rotate the array.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Rotate Array - LeetCode 189 Solution

Q: Which companies ask LeetCode 189. Rotate Array in interviews?

This problem is frequently asked at Accenture, American Express, Capgemini, Capital One, DXC Technology and 17 more companies.

Rotate Array - Complete Solution Guide

Rotate Array is LeetCode problem 189, 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

Given an integer array nums , rotate the array to the right by k steps, where k is non-negative. Example 1: Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Explanation: rotate 1 steps to the right: [7,1,2,3,4,5,6] rotate 2 steps to the right: [6,7,1,2,3,4,5] rotate 3 steps to the right: [5,6,7,1,2,3,4] Example 2: Input: nums = [-1,-100,3,99], k = 2 Output: [3,99,-1,-100] Explanation: rotate 1 steps to the right: [99,-1,-100,3] rotate 2 steps to the right: [3,99,-1,-100] Constraints:

Detailed Explanation

The problem requires rotating an array `nums` to the right by `k` steps. This means moving the last `k` elements to the beginning of the array and shifting the remaining elements to the right. The rotation should be done in-place, meaning we should modify the original array directly without using extra space proportional to the input size. The value of `k` can be larger than the array's length, so we need to handle that as well.

Solution Approach

The solution utilizes the reversal algorithm. First, it reduces 'k' using the modulo operator to ensure it's within the array's bounds. Then, it reverses the entire array. After that, it reverses the first 'k' elements and then reverses the remaining 'n-k' elements. This combination of reversals effectively rotates the array to the right by 'k' positions.

Step-by-Step Algorithm

Step 1: Calculate k % n to handle cases where k > n. This ensures k is within the valid range of array indices.
Step 2: Reverse the entire array nums from index 0 to n-1.
Step 3: Reverse the first k elements of the array nums from index 0 to k-1.
Step 4: Reverse the remaining n-k elements of the array nums from index k to n-1.

Key Insights

Insight 1: Using the modulo operator (%) is crucial to handle cases where 'k' is greater than the length of the array 'nums'. This ensures that the rotation happens within the bounds of the array.
Insight 2: The problem can be efficiently solved in-place by reversing array segments. Reversing the entire array, then reversing the first k elements, and finally reversing the remaining elements will effectively rotate the array.
Insight 3: Reversing is an O(n) operation, so performing a constant number of reverses results in an overall linear time complexity.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Two Pointers.

Companies

Asked at: Accenture, American Express, Capgemini, Capital One, DXC Technology, EPAM Systems, Flipkart, Goldman Sachs, IBM, Infosys, Netflix, Oracle, Samsung, Scale AI, ServiceNow, Visa, Walmart Labs, Wipro, Zoho, eBay, razorpay, tcs.

Rotate Array | LeetCode 189

Medium

Array Math Two Pointers

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

Solution Code

Available in:

class Solution {
    public void rotate(int[] nums, int k) {
        int n = nums.length;
        k %= n;

        function reverse = (start, end) -> {
            while (start < end) {
                int temp = nums[start];
                nums[start] = nums[end];
                nums[end] = temp;
                start++;
                end--;
            }
        };

        reverse.apply(0, n - 1);
        reverse.apply(0, k - 1);
        reverse.apply(k, n - 1);
    }

    interface function {
        void apply(int start, int end);
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 189: Rotate Array - Detailed explanation and solution approaches.

Topics

Array Math Two Pointers

Companies

AccentureAmerican ExpressCapgeminiCapital OneDXC TechnologyEPAM SystemsFlipkartGoldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:22

Problem #:189

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: Calculate k % n to handle cases where k > n. This ensures k is within the valid range of array indices.
2
Step 2: Reverse the entire array nums from index 0 to n-1.
3
Step 3: Reverse the first k elements of the array nums from index 0 to k-1.
4
Step 4: Reverse the remaining n-k elements of the array nums from index k to n-1.

💡

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 derived from the three reversal operations, each taking linear time. The space complexity is constant because we only use a few variables to track indices during the reversal process, regardless of the array 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

Using Extra Space

Create a new array of the same size, copy the last 'k' elements of the original array to the beginning of the new array, and then copy the remaining elements from the original array to the rest of the new array. Finally, copy the new array back into the original array.

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

Trade-offs: Simpler to understand initially, but uses extra space, violating the in-place requirement. It may be preferable if modifying the original array is not allowed.

Cyclic Replacements

Iterate through the array, performing a cyclic replacement of elements. Start at index 0, move the element at index 0 to index k, the element at index k to index 2k (modulo n), and so on, until you return to index 0. If you haven't visited all elements, start at index 1 and repeat. This avoids extra space.

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

Trade-offs: More complex to implement and reason about than the reversal method. The reversal method is more common and easier to remember.

✨

Key Insights

1Insight 1: Using the modulo operator (%) is crucial to handle cases where 'k' is greater than the length of the array 'nums'. This ensures that the rotation happens within the bounds of the array.
2Insight 2: The problem can be efficiently solved in-place by reversing array segments. Reversing the entire array, then reversing the first k elements, and finally reversing the remaining elements will effectively rotate the array.
3Insight 3: Reversing is an O(n) operation, so performing a constant number of reverses results in an overall linear time complexity.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Edge case 1: k = 0. The algorithm correctly handles this, as reversing the array three times results in no change to the original array.
Edge case 2: k = n (or a multiple of n). After the modulo operation, k becomes 0, and the algorithm behaves as expected, leaving the array unchanged.
Edge case 3: n = 1. The array contains only one element. The modulo operator will result in k=0 and the array stays unchanged which is correct.
Edge case 4: Empty Array nums = []. The code implicitly handles this due to the conditional `while start < end` in the reverse function, as start will never be less than end.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to use the modulo operator (%) when k is larger than the array's length. This can lead to out-of-bounds errors or incorrect rotation results.

✗Common mistake 2: Not handling the in-place requirement and creating a new array instead, which increases space complexity.

✗Common mistake 3: Incorrectly implementing the reversal logic, which can lead to elements being overwritten before they are rotated to their correct positions.

✅

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: Data encryption. Rotating data blocks is a common technique in cryptography.

🏦Real-world scenario 2: Circular buffers. Rotating arrays are used to implement circular buffers, which are useful for managing data streams where data is continuously overwritten in a loop.

🎮Real-world scenario 3: Image processing. Rotating image pixels can be accomplished using array rotation techniques.

💡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: June 10, 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

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

@AlgoMasterJun 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.

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 "Rotate Array"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?