#650

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

2 Keys Keyboard

Minimize the number of operations to get exactly 'n' characters on the screen using two operations: Copy All and Paste.

Problem Statement

Initially, there is only one character 'A' on the screen of a notepad. You can perform one of two operations for each step: 'Copy All' (which copies all characters currently on the screen to the clipboard) or 'Paste' (which adds all characters in the clipboard to the screen). Given an integer n, your task is to return the minimum number of operations required to get exactly n characters on the screen.

For example, if n = 3, you can perform the following steps: In the first step, you use 'Copy All'. In the second step, you use 'Paste' to get 'AA'. In the third step, you use 'Paste' again to get 'AAA'. This results in 3 operations. Your goal is to calculate the minimum number of operations for any given n.

Examples

Example 1

Input: n = 3

Output: 3

Initially, we have one character 'A'. In step 1, we use Copy All operation. In step 2, we use Paste operation to get 'AA'. In step 3, we use Paste operation to get 'AAA'.

Example 2

Input: n = 1

Output: 0

Example details omitted.

Constraints

1 <= n <= 1000

Solution Approach

State Transition Dynamic Programming

To minimize operations, treat each state of the notepad as a number of 'A's on the screen, with the goal of transitioning from 1 'A' to n 'A's. Use dynamic programming to track the minimum number of operations required to reach each state, considering both the 'Copy All' and 'Paste' operations. This approach avoids redundant calculations and ensures an optimal solution.

Mathematical Insight for Optimization

Recognize that the number of operations can be optimized by looking at factors of n. If n is divisible by a smaller number, a 'Copy All' followed by multiple 'Paste' operations is more efficient. Hence, the problem can be broken down into determining the factors of n and using dynamic programming to minimize steps based on these divisions.

Iterative Calculation with Time Complexity O(√n)

For each number from 1 to n, compute the minimal steps using its factors and store intermediate results. The algorithm primarily depends on the factorization of n, ensuring that the time complexity is O(√n) due to the factor search, leading to significant performance improvements over brute-force methods.

Complexity Analysis

Metric	Value
Time	O(\sqrt{n})
Space	O(1)

The time complexity of the solution is O(√n) because for each number from 1 to n, we need to consider its divisors, and there are at most √n divisors for any number. The space complexity is O(1) because the solution only requires a constant amount of space beyond the input.

What Interviewers Usually Probe

Candidate demonstrates a strong understanding of dynamic programming and optimization techniques.
Candidate applies mathematical insights, considering divisors and states, to minimize operations.
Candidate is able to explain how state transition and factorization techniques work together.

Common Pitfalls or Variants

Common pitfalls

Not considering the factorization of n and how it affects the number of operations.
Overcomplicating the problem with unnecessary calculations or inefficient approaches.
Failing to optimize for time complexity when implementing dynamic programming solutions.

Follow-up variants

Consider variations where other operations besides 'Copy All' and 'Paste' are allowed.
Extend the problem to handle cases with multiple characters initially on the screen.
Implement a solution for a larger range of n values, focusing on efficiency for large inputs.

FAQ

What is the best approach for solving the 2 Keys Keyboard problem?

The best approach is to use state transition dynamic programming, focusing on minimizing operations by considering factors of n.

How does the 'Copy All' operation impact the solution?

The 'Copy All' operation is crucial in optimizing the number of operations, as it allows for repeated pastes, which reduces the total operation count.

What is the time complexity of the optimal solution?

The time complexity is O(√n), as we only need to consider the divisors of n to minimize the number of operations.

Can the problem be solved using brute force?

While brute force is possible, it is inefficient. The optimal solution uses dynamic programming to significantly reduce redundant calculations.

What is the role of dynamic programming in solving this problem?

Dynamic programming helps store intermediate results, ensuring that redundant calculations are avoided, and that the solution is efficient.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def minSteps(self, n: int) -> int:
        @cache
        def dfs(n):
            if n == 1:
                return 0
            i, ans = 2, n
            while i * i <= n:
                if n % i == 0:
                    ans = min(ans, dfs(n // i) + i)
                i += 1
            return ans

        return dfs(n)

Solution 2

#### Python3

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class Solution:
    def minSteps(self, n: int) -> int:
        @cache
        def dfs(n):
            if n == 1:
                return 0
            i, ans = 2, n
            while i * i <= n:
                if n % i == 0:
                    ans = min(ans, dfs(n // i) + i)
                i += 1
            return ans

        return dfs(n)

Solution 3

#### Java

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class Solution:
    def minSteps(self, n: int) -> int:
        @cache
        def dfs(n):
            if n == 1:
                return 0
            i, ans = 2, n
            while i * i <= n:
                if n % i == 0:
                    ans = min(ans, dfs(n // i) + i)
                i += 1
            return ans

        return dfs(n)

Continue Practicing

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