2 Keys Keyboard

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;
    }
};

This summary explains a C++ solution for the "2 Keys Keyboard" problem, where the goal is to compute the minimum number of steps to create a given number of characters using only two possible operations: copy all and paste.

• Begin with a function calculateMinSteps which takes an integer number as input, representing the total number of characters to be formed.

• Initialize two integers, result to store the cumulative number of steps, and divisor starting at 2, to factorize the number.

• Use a nested while loop:

  • The outer while loop continues until number is greater than 1.
  • The inner while loop executes while number can be evenly divided by divisor. Each time this condition is met:
    • Add the value of divisor to result.
    • Divide number by divisor to reduce it.

• Increment divisor after the inner loop finishes, moving to the next potential factor.

• The function returns result as the final count of minimum steps needed to reach the target number of characters.

The approach leverages prime factorization, noticing that the least costly way to multiply the pasted characters is by repeatedly using the most efficient multiplication represented by the prime factors of the target number.

class Solution {

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

The provided Java solution addresses the problem of determining the minimum number of operations required to get n characters starting from the first character on a keyboard, using operations that allow copying and pasting. The core idea is to decompose the target number n into its prime factors. The strategy involves dividing n repeatedly by its smallest divisor (starting from 2) while adding that divisor to a cumulative result until n becomes 1.

Key highlights of the code:

• Initialize result to store the total count of operations.
• Use a while loop to handle the division of the number by its divisors, starting from 2.
• Nested within this, another while loop checks if the divisor divides number completely without remainder; in such cases, the divisor is added to result and the number is divided by the divisor.
• Increment divisor by 1 after exploring all multiple divisibility by the current divisor.
• Return the value of result which will provide the minimum operations needed.

This code effectively uses factorization to determine the most efficient way (minimum operations) to reach the desired number of characters via strategic copy-paste operations modeled in a mathematical way.

class Solution:
    def minimumSteps(self, number: int) -> int:
        result = 0
        divisor = 2
        while number > 1:
            while number % divisor == 0:
                result += divisor
                number //= divisor
            divisor += 1
        return result

The "2 Keys Keyboard" problem involves finding the minimum number of steps to produce a specific number of 'A's on a screen using only two operations: copy all and paste. The solution provided uses Python to functionally determine the minimum steps.

Here’s how the solution works:

1. Initialize the result variable to zero to accumulate the total number of operations.
2. Start with the initial divisor value of 2, which signifies the first possible factorization other than 1.
3. Use a while loop to check if the input number is greater than 1.
4. Inside this while loop, the inner loop checks if the number is divisible by the current divisor without a remainder.
5. If so, the result accumulates the divisor's value (simulating the operations needed), and the author divides the number by the divisor (simulating the reduction of problem size).
6. Increment the divisor after processing to check the next possible factor.
7. The process repeats until the number cannot be reduced further, implying the optimal number of operations is achieved.

Return result, which holds the minimized operation count needed. The use of factorization here cleverly reduces the problem size by breaking down the number into its constituent factors, optimizing the number of operations needed compared to if trying to handle it count incrementally from 1. This approach efficiently addresses the problem in logarithmic time complexity related to the number's prime factors.