2 Keys Keyboard

Problem Statement

In this problem, we start with a single occurrence of the character 'A' displayed on a notepad. To manipulate this notepad, we have two operations at our disposal:

1. Copy All: This operation allows you to copy all characters currently visible on the screen. It must be noted that only a full copy of all characters is allowed, not just a partial selection.
2. Paste: This operation pastes the last copied set of characters onto the screen.

Given a target integer n, the challenge is to find the minimum number of operations required to have n instances of 'A' on the screen. The operations should be optimally chosen to meet the goal with as few steps as possible.

Examples

Example 1

Input:

n = 3

Output:

3

Explanation:

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

Constraints

• 1 <= n <= 1000

Approach and Intuition

To understand the minimum number of operations required to replicate 'A' on the notepad to reach exactly n iterations, we can examine a few scenarios and develop an intuition based on the constraints:

1. If n is 1, you start with one 'A'. No operations are needed since you're already at the desired state.

2. If n is greater than 1, explore the factorization of n:

   • Every factor of n represents a potential breakpoint where a Copy All + Paste operation sequence might be optimized. For instance, if n is 6, one of the optimal ways is to get to 3 'A's through minimal operations and then just duplicate that twice.
   • Specifically, for any factor f of n where f is less than n, one can split this into replicating f 'A's and then using the operations to double, triple, etc., until reaching n by multiplying f. The number of operations this requires corresponds to the number of times f must be pasted (which is n / f) plus the operations needed to reach f initially.

3. To determine the fewest operations, iterate through potential factors and determine the sequence size based often on smaller sub-problems — this aligns closely with dynamic programming or a greedy approach where at each step, the problem is broken down into smaller sub-problems.

The essence of the solution is derived from the fact that efficient replication (via pasting) hinges on the factors of n. Voyager into deeper, endeavor into efficient algorithm design perhaps could look into progressively building up the number of 'A's and utilizing each step most advantageous, considering the number operations it necessitates.

Solutions

class Solution {
public:
    int calculateMinSteps(int number) {
        int result = 0;
        int divisor = 2;
        while (number > 1) {
            while (number % divisor == 0) {
                result += divisor;
                number /= divisor;
            }
            divisor++;
        }
        return result;
    }
};