onenoughtone

Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

Dynamic Programming Approach - Complex approach
Prime Factorization Approach - Complex approach
Optimized Dynamic Programming Approach - Complex approach

Approach 1: Dynamic Programming Approach

The key insight for solving this problem is to use dynamic programming to find the minimum number of operations needed for each number of 'A's from 1 to n.

Let's define dp[i] as the minimum number of operations to get exactly i 'A's on the screen. We know that dp[1] = 0 since we start with one 'A'.

For each i from 2 to n, we need to find the minimum number of operations. We can achieve this by considering all possible ways to reach i 'A's by copying and pasting from a smaller number of 'A's.

Specifically, for each j that is a divisor of i (where j < i), we can copy all j 'A's and then paste (i/j) - 1 times. This would take a total of dp[j] + (i/j) operations (dp[j] to get to j 'A's, 1 operation to copy all, and (i/j) - 1 operations to paste).

We choose the minimum of these values for dp[i].

Algorithm:

Initialize an array dp of size n+1, where dp[i] represents the minimum number of operations to get i 'A's.
Set dp[1] = 0, since we start with one 'A'.
For each i from 2 to n:
a. Initialize dp[i] to a large value (e.g., infinity).
b. For each j from 1 to i-1:
i. If i is divisible by j, update dp[i] = min(dp[i], dp[j] + (i/j)).
Return dp[n] as the minimum number of operations.

Time Complexity:

O(n²)

We need to compute dp[i] for each i from 1 to n, and for each i, we consider all possible j from 1 to i-1.

Space Complexity:

O(n)

We use an array of size n+1 to store the minimum number of operations for each number of 'A's.

Approach 2: Prime Factorization Approach

An alternative approach is to use the observation that the minimum number of operations is the sum of the prime factors of n (counting duplicates).

This is because the most efficient way to get n 'A's is to build up the number by multiplying by its prime factors. For each prime factor p, we need p operations: 1 to copy and p-1 to paste.

For example, if n = 6 = 2 × 3, we need 2 + 3 = 5 operations:

Start with 'A'
Copy All (clipboard has 'A')
Paste (screen has 'AA')
Copy All (clipboard has 'AA')
Paste (screen has 'AAAA')
Paste (screen has 'AAAAAA')

Note that we only count the operations from step 2 onwards, so that's 5 operations.

Algorithm:

Initialize a variable result to 0.
For each potential prime factor i from 2 to sqrt(n):
a. While n is divisible by i:
i. Add i to result.
ii. Divide n by i.
If n > 1 (n is a prime number greater than sqrt(n)), add n to result.
Return result as the minimum number of operations.

Time Complexity:

O(sqrt(n))

We only need to check potential prime factors up to the square root of n.

Space Complexity:

O(1)

We only use a constant amount of extra space regardless of the input size.

Approach 3: Optimized Dynamic Programming Approach

We can optimize the dynamic programming approach by only considering the divisors of i when calculating dp[i], rather than checking all j from 1 to i-1.

This is because if j is not a divisor of i, we cannot get exactly i 'A's by copying all j 'A's and then pasting.

This optimization reduces the time complexity from O(n²) to O(n * sqrt(n)), as the number of divisors of a number is approximately O(sqrt(n)).

Algorithm:

Initialize an array dp of size n+1, where dp[i] represents the minimum number of operations to get i 'A's.
Set dp[1] = 0, since we start with one 'A'.
For each i from 2 to n:
a. Initialize dp[i] to a large value (e.g., infinity).
b. For each j from 1 to sqrt(i):
i. If i is divisible by j, update dp[i] = min(dp[i], dp[j] + (i/j)).
ii. If j is not equal to i/j and i is divisible by j, update dp[i] = min(dp[i], dp[i/j] + j).
Return dp[n] as the minimum number of operations.

Time Complexity:

O(n * sqrt(n))

We need to compute dp[i] for each i from 1 to n, and for each i, we consider all divisors of i, which is approximately O(sqrt(n)).

Space Complexity:

O(n)

We use an array of size n+1 to store the minimum number of operations for each number of 'A's.

Pseudocode

solution.pseudo

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function minSteps(n):
    // Initialize dp array
    dp = new array of size n+1
    dp[1] = 0  // We start with one &apos;A&apos;
 
    for i from 2 to n:
        dp[i] = infinity
        for j from 1 to i-1:
            if i is divisible by j:
                dp[i] = min(dp[i], dp[j] + (i/j))
 
    return dp[n]

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