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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.
#include <vector>
#include <climits>
int minSteps(int n) {
if (n == 1) return 0;
std::vector<int> dp(n + 1, INT_MAX);
dp[1] = 0;
for (int i = 2; i <= n; ++i) {
for (int j = 1; j < i; ++j) {
if (i % j == 0) {
dp[i] = std::min(dp[i], dp[j] + i / j);
}
}
}
return dp[n];
}
int main() {
int n = 3;
std::cout << "Minimum steps to get " << n << " A's: " << minSteps(n) << std::endl;
return 0;
}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.
1function minSteps(n) {
2 let count = 0, d = 2;
3 while (n > 1) {
4 while (n % d === 0) {
5 count += d;
6 n /= d;
7 }
8 d++;
9 }
10 return count;
11}
12
13const n = 3;
14console.log(`Minimum steps to get ${n} A's: ${minSteps(n)}`);The solution written in JavaScript follows a similar logic of decomposing n into its prime factors and accumulating these factors to determine minimum operations.
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