Problem: 2 Keys Keyboard

The problem asks for the minimum number of operations (copy all and paste) to get 'A' exactly n times on a screen, starting with one 'A'.

Solutions Explained

Several approaches solve this problem:

Solution 1: Top-Down Dynamic Programming with Memoization

This solution uses recursion with memoization to avoid redundant calculations.

Approach:

The core idea is that to obtain n 'A's, you might have copied k 'A's and pasted them n/k times. The minimum steps will be the steps to get k 'A's plus n/k (the number of pastes). We explore all possible divisors k of n. Memoization stores previously computed results for different values of n to significantly speed up the process.

Time Complexity: O(√n) - The time complexity is dominated by the loop that iterates through divisors of n. In the worst case, we check up to √n divisors (because divisors come in pairs). The memoization ensures that each subproblem (dfs(n/k)) is solved only once.

Space Complexity: O(n) - The space complexity is due to the memoization table (f or cache). In the worst case, the table could store results for all numbers up to n.

Code (Python):

class Solution:
    def minSteps(self, n: int) -> int:
        @cache # This is a Python decorator for memoization
        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)

Other languages (Java, C++, Go) follow a similar structure, using their respective memoization techniques (arrays or maps).

Solution 2: Bottom-Up Dynamic Programming

This approach uses iterative dynamic programming to build a solution from the ground up.

Approach:

We create a dp array where dp[i] stores the minimum steps to get i 'A's. We initialize dp[1] = 0 (no steps needed for one 'A'). Then, we iterate from 2 to n. For each i, we consider all divisors j of i. The minimum steps to get i 'A's will be the minimum of dp[i/j] + j, where dp[i/j] is the steps to get i/j 'A's and j represents the copy-paste operation.

Time Complexity: O(n√n) - The outer loop iterates n times. The inner loop iterates up to √i times for each i. The total time complexity becomes roughly O(n√n).

Space Complexity: O(n) - The space complexity is due to the dp array of size n+1.

Code (Python):

class Solution:
    def minSteps(self, n: int) -> int:
        dp = list(range(n + 1))
        dp[1] = 0
        for i in range(2, n + 1):
            j = 2
            while j * j <= i:
                if i % j == 0:
                    dp[i] = min(dp[i], dp[i // j] + j)
                j += 1
        return dp[-1]

Other languages use similar iterative dynamic programming approaches.

Solution 3: Prime Factorization

This is the most efficient solution, leveraging the mathematical properties of the problem.

Approach:

The problem can be viewed as a prime factorization problem. To obtain n 'A's, we repeatedly copy and paste. The number of steps corresponds to the sum of the prime factors of n (with multiplicities). For example, if n = 12 = 2^2 * 3, the minimum steps are 2 + 2 + 3 = 7 (copy 2, paste, copy 2, paste, copy 3, paste).

Time Complexity: O(√n) - The time complexity is dominated by the loop that iterates up to the square root of n to find prime factors.

Space Complexity: O(1) - The algorithm uses a constant amount of extra space.

Code (Java):

class Solution {
    public int minSteps(int n) {
        int res = 0;
        for (int i = 2; n > 1; ++i) {
            while (n % i == 0) {
                res += i;
                n /= i;
            }
        }
        return res;
    }
}

Choosing the Best Solution:

Solution 3 (prime factorization) is the most efficient in terms of time complexity, making it the preferred approach for large values of n. Solutions 1 and 2 are useful for understanding dynamic programming techniques but are less efficient.