What is the time complexity of 2 Keys Keyboard?

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

What data structures are used to solve 2 Keys Keyboard?

This problem uses Math, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving 2 Keys Keyboard?

Insight 1: The number of operations needed to reach 'n' is directly related to the prime factorization of 'n'. Each prime factor corresponds to a 'copy' and then a series of 'pastes'. Insight 2: If 'n' is divisible by 'x', it means we can reach 'n' by copying 'n/x' 'A's and then pasting 'x-1' times (plus the initial copy operation). The total operations are 1 + (x-1) = x (to reach 'n/x' 'A's there will be operations required).

Which companies ask LeetCode 650. 2 Keys Keyboard in interviews?

This problem is frequently asked at Salesforce.

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2 Keys Keyboard - LeetCode 650 Solution

2 Keys Keyboard - Complete Solution Guide

2 Keys Keyboard is LeetCode problem 650, 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 only one character 'A' on the screen of a notepad. You can perform one of two operations on this notepad for each step: Copy All: You can copy all the characters present on the screen (a partial copy is not allowed). Paste: You can paste the characters which are copied last time. Given an integer n , return the minimum number of operations to get the character 'A' exactly n times on the screen . Example 1: Input: n = 3 Output: 3 Explanation: Initially, we have one character 'A'. In step

Detailed Explanation

The problem asks us to find the minimum number of operations required to obtain exactly 'n' characters 'A' on a notepad, starting with only one 'A'. The allowed operations are 'Copy All' (copy the entire content of the notepad) and 'Paste' (paste the last copied content). We need to determine the fewest Copy All and Paste operations to reach 'n' 'A's.

Solution Approach

The solution utilizes prime factorization. It iteratively finds the smallest prime factor of 'n'. Each time a prime factor 'd' is found, it means we performed 'd' operations (1 Copy All and d-1 Paste to multiply the current number of 'A's by 'd'). The process repeats until 'n' becomes 1. The sum of all prime factors gives the minimum number of operations. If 'n' remains greater than 1 after the loop, it signifies that 'n' itself is a prime number, and we need to add 'n' to the total operations.

Step-by-Step Algorithm

Step 1: Handle the base case where n = 1. If n is 1, return 0 since no operations are needed.
Step 2: Initialize 'ans' to 0, which will store the total number of operations.
Step 3: Initialize 'd' to 2, which will be used to find the smallest prime factor of 'n'.
Step 4: Iterate while 'd * d <= n'. This condition is used because if 'n' has a prime factor greater than the square root of 'n', it must also have a prime factor smaller than the square root of 'n'. Thus we only need to check divisibility up to the square root of n.
Step 5: While 'n' is divisible by 'd', add 'd' to 'ans' (representing the 'd' operations to obtain 'n' from 'n/d' 'A's). Then, divide 'n' by 'd' to update 'n'.
Step 6: Increment 'd' to check for the next potential prime factor.
Step 7: After the loop, if 'n' is still greater than 1, it means 'n' is a prime number itself. Add 'n' to 'ans'.
Step 8: Return 'ans', which represents the minimum number of operations.

Key Insights

Insight 1: The number of operations needed to reach 'n' is directly related to the prime factorization of 'n'. Each prime factor corresponds to a 'copy' and then a series of 'pastes'.
Insight 2: If 'n' is divisible by 'x', it means we can reach 'n' by copying 'n/x' 'A's and then pasting 'x-1' times (plus the initial copy operation). The total operations are 1 + (x-1) = x (to reach 'n/x' 'A's there will be operations required).
Insight 3: Finding the smallest prime factors first leads to the optimal (minimum) number of operations. This is because smaller factors require fewer paste operations.

Complexity Analysis

Time Complexity: O(sqrt(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming.

Companies

Asked at: Salesforce.

2 Keys Keyboard | LeetCode 650

Medium

Math Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int minSteps(int n) {
        if (n == 1) {
            return 0;
        }
        int ans = 0;
        int d = 2;
        while (d * d <= n) {
            while (n % d == 0) {
                ans += d;
                n /= d;
            }
            d += 1;
        }
        if (n > 1) {
            ans += n;
        }
        return ans;
    }
}

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Problem Info

Description

Problem 650: 2 Keys Keyboard - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming

Companies

Salesforce

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:650

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: Handle the base case where n = 1. If n is 1, return 0 since no operations are needed.
2
Step 2: Initialize 'ans' to 0, which will store the total number of operations.
3
Step 3: Initialize 'd' to 2, which will be used to find the smallest prime factor of 'n'.
4
Step 4: Iterate while 'd * d <= n'. This condition is used because if 'n' has a prime factor greater than the square root of 'n', it must also have a prime factor smaller than the square root of 'n'. Thus we only need to check divisibility up to the square root of n.
5
Step 5: While 'n' is divisible by 'd', add 'd' to 'ans' (representing the 'd' operations to obtain 'n' from 'n/d' 'A's). Then, divide 'n' by 'd' to update 'n'.
6
Step 6: Increment 'd' to check for the next potential prime factor.
7
Step 7: After the loop, if 'n' is still greater than 1, it means 'n' is a prime number itself. Add 'n' to 'ans'.
8
Step 8: Return 'ans', which represents the minimum number of operations.

💡

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(sqrt(n))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(sqrt(n)) because the outer loop iterates up to the square root of 'n'. The inner loop executes multiple times, but in total, the number of divisions performed in the inner loop is limited by the number of prime factors of 'n'. The space complexity is O(1) because the algorithm uses a fixed number of variables regardless of the input 'n'.

📖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

Dynamic Programming

A dynamic programming approach could be used where dp[i] represents the minimum steps to get 'i' A's. However, it's less efficient.

Time: O(n^2) or O(n log n) depending on the specific implementation.Space: O(n)

Trade-offs: DP is more intuitive for some, but less efficient due to its higher time and space complexity. The provided solution (prime factorization) is significantly faster, especially for larger values of 'n'.

✨

Key Insights

1Insight 1: The number of operations needed to reach 'n' is directly related to the prime factorization of 'n'. Each prime factor corresponds to a 'copy' and then a series of 'pastes'.
2Insight 2: If 'n' is divisible by 'x', it means we can reach 'n' by copying 'n/x' 'A's and then pasting 'x-1' times (plus the initial copy operation). The total operations are 1 + (x-1) = x (to reach 'n/x' 'A's there will be operations required).
3Insight 3: Finding the smallest prime factors first leads to the optimal (minimum) number of operations. This is because smaller factors require fewer paste operations.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The solution correctly handles this case by returning 0, as no operations are needed.
Edge case 2: n is a prime number. The solution correctly identifies this after the while loop (d * d <= n) and adds n to the answer.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider the case where 'n' is a prime number. After the loop, if 'n' > 1, it needs to be added to the result.

✗Common mistake 2: Trying to iterate through all numbers up to 'n' instead of the square root of 'n'. This leads to a much higher time complexity.

✅

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: Compression algorithms where finding repeating patterns and minimizing operations is crucial for efficient compression.

🏦Real-world scenario 2: Optimization problems where minimizing the number of steps to achieve a desired state is important, like in resource allocation or task scheduling.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: September 19, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "2 Keys Keyboard"?

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

🤖 Need help with this problem?