How do you solve 2 Keys Keyboard on LeetCode?

2 Keys Keyboard (LeetCode #650) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the relationship between operations and factors of n is crucial.

What is the brute force solution for 2 Keys Keyboard?

The brute force approach for 2 Keys Keyboard is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating every possible sequence of operations to reach 'A' n times. This means trying every combination of Copy and Paste operations, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve 2 Keys Keyboard?

2 Keys Keyboard uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Factorization, Dynamic Programming.

What are the interview tips for 2 Keys Keyboard?

Always think about how you can break down the problem into smaller parts. Consider edge cases, like n = 1, early in your solution. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving 2 Keys Keyboard?

Forgetting to check for factors correctly. Not considering the optimal sequence of operations.

#650

2 Keys Keyboard

Medium

MathDynamic ProgrammingFactorizationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the idea of factorization. Instead of simulating every operation, we can find the minimum operations by considering how many times we can copy and paste based on the factors of n.

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Algorithm

4 steps

1Step 1: Initialize operations to 0.
2Step 2: For each integer i from 2 to n, check if i is a factor of n.
3Step 3: If it is, add i to operations and divide n by i.
4Step 4: Repeat until n becomes 1.

solution.py10 lines

1def minOperations(n):
2    operations = 0
3    i = 2
4    while n > 1:
5        if n % i == 0:
6            operations += i
7            n //= i
8        else:
9            i += 1
10    return operations

ℹ

Complexity note: The time complexity is O(n) because we iterate through potential factors up to n, and for each factor, we may divide n multiple times.

1Understanding the relationship between operations and factors of n is crucial.
2Each operation can be viewed as a way to build up to n using its factors.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.