Problem Description

You start with a notepad that has a single character 'A' on the screen. You can perform two operations:

1. Copy All: Copy all characters currently on the screen (you cannot copy just a portion)
2. Paste: Paste the characters that were most recently copied

Your goal is to get exactly n 'A' characters on the screen using the minimum number of operations.

For example, if n = 3:

• Start with: A (1 character)
• Operation 1: Copy All (clipboard now has A)
• Operation 2: Paste (screen now has AA, 2 characters)
• Operation 3: Paste (screen now has AAA, 3 characters)
• Total: 3 operations

The solution uses dynamic programming with memoization to find the minimum operations. The key insight is that to reach n characters, we can factorize n and use its divisors. If n is divisible by i, we can first get n/i characters, then copy once and paste i-1 times (total i operations) to multiply the count by i.

The recursive function dfs(n) returns the minimum operations to get n characters:

• Base case: If n = 1, no operations needed (return 0)
• For each divisor i of n, we can get n/i characters first, then perform i operations (1 copy + i-1 pastes) to reach n
• The answer is the minimum among all possible paths

Intuition

The key observation is that reaching n characters is fundamentally about multiplication. When we copy and paste, we're essentially multiplying the current number of characters.

Think about how we get from 1 to n characters. Every sequence of operations can be broken down into groups where each group consists of one Copy followed by several Pastes. If we Copy when we have x characters and then Paste k times, we end up with x * (k+1) characters.

This means to reach n, we need to find a way to express n as a product of factors. For instance, to get 12 characters, we could:

• Get to 3 characters, then Copy + Paste 3 times (multiply by 4): 3 × 4 = 12
• Get to 4 characters, then Copy + Paste 2 times (multiply by 3): 4 × 3 = 12
• Get to 6 characters, then Copy + Paste 1 time (multiply by 2): 6 × 2 = 12

The number of operations for each multiplication by factor f is exactly f (1 Copy + f-1 Pastes). This leads us to think about prime factorization. If n = p1 × p2 × p3..., then the minimum operations would be p1 + p2 + p3 + ...

However, we don't always want to use prime factors. Sometimes grouping primes together gives better results. For example, for n = 6 = 2 × 3, we could use 2 + 3 = 5 operations, but we could also treat 6 as 6 × 1, though that would need 6 operations.

This naturally suggests a recursive approach: for each divisor i of n, we can first get to n/i characters (recursively), then multiply by i using i operations. We try all divisors and pick the minimum. The problem decomposes into smaller subproblems, making dynamic programming with memoization an ideal solution strategy.

Learn more about Math and Dynamic Programming patterns.

Solution Approach

The solution implements a top-down dynamic programming approach with memoization using Python's @cache decorator.

Implementation Details:

1. Recursive Function dfs(n): This function calculates the minimum operations needed to get exactly n characters.

2. Base Case: When n = 1, we already have one 'A' on the screen, so no operations are needed. Return 0.

3. Recursive Case: For any n > 1, we iterate through all possible divisors from 2 to √n:

   • Start with i = 2 and initialize ans = n (worst case: copy once, then paste n-1 times)
   • Check if i is a divisor of n (i.e., n % i == 0)
   • If yes, we can:
     • First get n // i characters (recursive call: dfs(n // i))
     • Then multiply by i using i operations (1 copy + i-1 pastes)
     • Total operations: dfs(n // i) + i
   • Keep track of the minimum among all valid divisors

4. Optimization: We only check divisors up to √n because:

   • If i is a divisor, then n/i is also a divisor
   • One of them must be ≤ √n
   • The recursive call dfs(n // i) will handle the larger divisors

5. Memoization: The @cache decorator automatically stores results of previous computations, preventing redundant calculations for the same value of n.

Example Walkthrough for n = 6:

• dfs(6) checks divisors: 2 and 3
• For divisor 2: dfs(3) + 2 = 3 + 2 = 5
• For divisor 3: dfs(2) + 3 = 2 + 3 = 5
• Both paths give 5 operations, which is optimal

The algorithm effectively finds the optimal factorization of n that minimizes the sum of factors, leveraging dynamic programming to avoid recomputing subproblems.

Example Walkthrough

Let's trace through the solution for n = 8 to understand how the algorithm finds the minimum operations.

Starting with n = 8:

We need to get from 1 'A' to 8 'A's. The algorithm explores different factorization paths:

1. First, dfs(8) is called:

   • Initialize ans = 8 (worst case: copy once, paste 7 times)
   • Check divisor i = 2: Since 8 % 2 = 0, it's valid
     • We can get to 8 by first reaching 4 characters, then multiplying by 2
     • Cost: dfs(4) + 2
   • Check divisor i = 3: 8 % 3 ≠ 0, skip
   • Check divisor i = 4: Since 8 % 4 = 0, it's valid
     • We can get to 8 by first reaching 2 characters, then multiplying by 4
     • Cost: dfs(2) + 4

2. Computing dfs(4):

   • Initialize ans = 4 (worst case)
   • Check divisor i = 2: Since 4 % 2 = 0, it's valid
     • Cost: dfs(2) + 2

3. Computing dfs(2):

   • Initialize ans = 2 (worst case)
   • No divisors to check (since we start from 2)
   • Return 2 (copy once, paste once)

4. Backtracking with computed values:

   • dfs(2) = 2
   • dfs(4) = dfs(2) + 2 = 2 + 2 = 4
   • dfs(8) compares:
     • Path via divisor 2: dfs(4) + 2 = 4 + 2 = 6
     • Path via divisor 4: dfs(2) + 4 = 2 + 4 = 6
   • Both give 6 operations (optimal)

Verifying the solution (6 operations):

• Start: A (1 character)
• Op 1: Copy All → clipboard: A
• Op 2: Paste → AA (2 characters)
• Op 3: Copy All → clipboard: AA
• Op 4: Paste → AAAA (4 characters)
• Op 5: Copy All → clipboard: AAAA
• Op 6: Paste → AAAAAAAA (8 characters)

The algorithm found that 8 = 2 × 2 × 2, requiring 2 + 2 + 2 = 6 operations, which corresponds to the sum of prime factors.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n^(3/2))

The algorithm uses memoized recursion to find the minimum steps. For each number n, it checks all potential divisors from 2 to √n.

• The while loop runs O(√n) times for each recursive call
• In the worst case (when n is prime), the recursion depth is O(log n)
• Due to memoization with @cache, each unique value from 1 to n is computed only once
• The total number of unique subproblems is at most O(n)
• For each subproblem, we spend O(√n) time checking divisors
• Therefore, the overall time complexity is O(n × √n) = O(n^(3/2))

Space Complexity: O(n)

The space complexity consists of:

• The recursion call stack depth: O(log n) in the worst case (when repeatedly dividing by smallest prime factors)
• The memoization cache storing results: O(n) in the worst case (could store results for all values from 1 to n)
• Since O(n) > O(log n), the overall space complexity is O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Understanding of Operations

A common mistake is misunderstanding how the copy-paste operations work. Some might think:

• Each paste operation adds 1 character (incorrect)
• You can selectively copy portions of the screen (incorrect)

Reality: When you paste, you add ALL characters from the clipboard. If you have "AA" on screen and copy it, then paste, you get "AAAA" (doubling), not "AAA".

Solution: Remember that copy-paste operations multiply the current count. To get from k characters to n characters where n = k * m, you need 1 copy + (m-1) pastes = m total operations.

2. Not Considering All Divisors Properly

The code only checks divisors up to √n, which is correct, but developers might forget why this works:

# Incorrect approach - checking all numbers up to n
for factor in range(2, current_n + 1):
    if current_n % factor == 0:
        min_steps = min(min_steps, dfs(current_n // factor) + factor)

This would be inefficient and redundant because when we find divisor i, the recursive call dfs(n // i) will handle the complementary divisor.

Solution: Trust the optimization - checking up to √n is sufficient because the recursive structure handles larger divisors through smaller subproblems.

3. Forgetting the Base Case or Setting It Incorrectly

Some might set the base case as:

if current_n == 1:
    return 1  # Wrong! We start with 1 'A' already

Solution: The base case should return 0 because we start with one 'A' on the screen - no operations needed to have 1 character.

4. Integer Overflow in Other Languages

While Python handles large integers automatically, in languages like Java or C++, calculating operations for large values of n might cause overflow:

// In Java, this might overflow for large n
int minSteps = n;  // If n is close to Integer.MAX_VALUE

Solution: Use appropriate data types (long in Java, long long in C++) or add bounds checking.

5. Inefficient Iteration Through Factors

Some might try to optimize by skipping even numbers after checking 2:

# Attempting to optimize but creating bugs
factor = 2
while factor * factor <= current_n:
    if current_n % factor == 0:
        min_steps = min(min_steps, dfs(current_n // factor) + factor)
    factor = 1 if factor == 2 else 2  # Bug! Should be factor += 1 or factor += 2

Solution: Keep the iteration simple with factor += 1. The performance gain from skipping even numbers is minimal compared to the memoization benefit, and it reduces the chance of off-by-one errors.