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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.
1#include <stdio.h>
2#include <limits.h>
3
4int minSteps(int n) {
5 if (n == 1) return 0;
6 int dp[n + 1];
7 dp[1] = 0;
8 for (int i = 2; i <= n; ++i) {
9 dp[i] = INT_MAX;
10 for (int j = 1; j < i; ++j) {
11 if (i % j == 0) {
12 dp[i] = dp[i] < dp[j] + i / j ? dp[i] : dp[j] + i / j;
13 }
14 }
15 }
16 return dp[n];
17}
18
19int main() {
20 int n = 3;
21 printf("Minimum steps to get %d A's: %d\n", n, minSteps(n));
22 return 0;
23}This C solution initializes an array of size n+1 to track minimum steps for each possible number of 'A's up to n. We iterate over each possible number of 'A's, checking for each j (where 1 <= j < i) if i can be entirely divided by j. If so, update dp[i] using dp[j] plus the number of pastes needed to reach i from j.
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.
This Java program leverages prime factorization by iterating through divisors and using each to shrink n. The sum of these factors gives the smallest number of operations.