2 Keys Keyboard

Description

There is only one character 'A' on the screen of a notepad. You can perform one of two operations on this notepad for each step:

• Copy All: You can copy all the characters present on the screen (a partial copy is not allowed).
• Paste: You can paste the characters which are copied last time.

Given an integer n, return the minimum number of operations to get the character 'A' exactly n times on the screen.

Example 1:

Input: n = 3 Output: 3 Explanation: Initially, we have one character ‘A’. In step 1, we use Copy All operation. In step 2, we use Paste operation to get ‘AA’. In step 3, we use Paste operation to get ‘AAA’.

Example 2:

Input: n = 1 Output: 0

Constraints:

• 1 <= n <= 1000

Code

Top Down DP with memoization

Time Complexity: $O(n^2)$, Space Complexity: $O(n)$

關鍵在於 $T(n)=T(i)+\frac{n}{i},\forall i\in 1\cdots ,n-1$, and n mod i == 0.

$\frac{n}{i}$ 是計算由 $i$ 個 $A$ Copy and Paste 到 $n$ 個 $A$ 所需要的 steps 次數，由 $\frac{n-i}{i}+1$ 組成，其中 $\frac{n-i}{i}$ 是 paste 的次數，$1$ 則是 copy 的次數。

class Solution {
public:
    int minSteps(int n) {
        vector<int> memo(n + 1, -1);
        return solve(n, memo);
    }
 
    int solve(int n, vector<int>& memo) {
        if(n == 1) return 0;
        if(memo[n] != -1) return memo[n];
        int res = INT_MAX;
        for(int i = 1; i < n; i++) {
            if(n % i != 0) continue;
            res = min(res, (n / i) + minSteps(i));
        }
        return memo[n] = res;
    }
};

Bottom Up DP

Time Complexity: $O(n^2)$, Space Complexity: $O(n)$

class Solution {
public:
    int minSteps(int n) {
        vector<int> dp(n + 1, 1001);
        dp[1] = 0;
 
        for(int i = 2; i <= n; i++) {
            for(int j = 1; j < i; j++) {
                if(i % j != 0) continue;
                dp[i] = min(dp[i], (i / j) + dp[j]);
            }
        }
        return dp[n];
    }
 
};

Source

• 2 Keys Keyboard - LeetCode