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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 <iostream>
2#include <vector>
3#include <climits>
4
5int minSteps(int n) {
6 if (n == 1) return 0;
7 std::vector<int> dp(n + 1, INT_MAX);
8 dp[1] = 0;
9 for (int i = 2; i <= n; ++i) {
10 for (int j = 1; j < i; ++j) {
11 if (i % j == 0) {
12 dp[i] = std::min(dp[i], dp[j] + i / j);
13 }
14 }
15 }
16 return dp[n];
17}
18
19int main() {
20 int n = 3;
21 std::cout << "Minimum steps to get " << n << " A's: " << minSteps(n) << std::endl;
22 return 0;
23}This C++ solution follows the same dynamic programming strategy as the C solution, employing a vector to store min steps for each count of 'A'. We update our dp vector for each factor of i to ensure minimized operations.
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.
public class Solution {
public int MinSteps(int n) {
int count = 0, d = 2;
while (n > 1) {
while (n % d == 0) {
count += d;
n /= d;
}
d++;
}
return count;
}
public static void Main(string[] args) {
Solution solution = new Solution();
int n = 3;
Console.WriteLine("Minimum steps to get " + n + " A's: " + solution.MinSteps(n));
}
}This C# method uses a factorization approach to compute the needed operations. It updates total operation count by summing divisors as n reduces by its smallest factors.