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This approach involves using a dynamic programming array where dp[i] represents the minimum number of operations required to achieve exactly i 'A's on the screen. The idea is to divide the problem into subproblems and solve each subproblem once to build up a solution for the original problem. For each integer i, if it can be obtained from j by first copying all from j and pasting (i/j-1) times, then dp[i] can be updated as dp[j] + i/j steps.
Time Complexity: O(n^2) due to the nested loops for divisor checking.
Space Complexity: O(n) to store the dynamic programming array.
1public class Solution {
2 public int minSteps(int n) {
3 if (n == 1) return 0;
4 int[] dp = new int[n + 1];
5 dp[1] = 0;
6 for (int i = 2; i <= n; i++) {
7 dp[i] = Integer.MAX_VALUE;
8 for (int j = 1; j < i; j++) {
9 if (i % j == 0) {
10 dp[i] = Math.min(dp[i], dp[j] + i / j);
11 }
12 }
13 }
14 return dp[n];
15 }
16
17 public static void main(String[] args) {
18 Solution solution = new Solution();
19 int n = 3;
20 System.out.println("Minimum steps to get " + n + " A's: " + solution.minSteps(n));
21 }
22}This Java solution implements the dynamic programming approach by maintaining an array where each index i represents minimum steps to build i 'A's. For each i, calculate its factors and use them to derive minimum steps.
The optimal strategy is based on a mathematical observation wherein the solution boils down to the prime factorization of n. Essentially, if you can find all factors of numbers up to n, the minimum operations needed are the sum of the prime factors of these numbers. Whenever you can make a sequence by repeating operations, using a simple Copy All and Paste enough times to reach a bigger number allows the number of operations to be minimized via multiplying smaller sequences.
Time Complexity: O(√n) as it iterates over possible factors up to the square root of n.
Space Complexity: O(1) with constant auxiliary space usage.
int minSteps(int n) {
int count = 0, d = 2;
while (n > 1) {
while (n % d == 0) {
count += d;
n /= d;
}
d++;
}
return count;
}
int main() {
int n = 3;
std::cout << "Minimum steps to get " << n << " A's: " << minSteps(n) << std::endl;
return 0;
}The C++ solution focuses on prime factor decomposition of n. By iteratively dividing n by the smallest factor, it aggregates the total operations needed, corresponding to sum of each factor discovered.