Leetcode 650.只有两个键的键盘

只有两个键的键盘

最初在一个记事本上只有一个字符 'A'。你每次可以对这个记事本进行两种操作：

1. Copy All (复制全部) : 你可以复制这个记事本中的所有字符(部分的复制是不允许的)。
2. Paste (粘贴) : 你可以粘贴你上一次复制的字符。

给定一个数字 n 。你需要使用最少的操作次数，在记事本中打印出恰好 n 个 'A'。输出能够打印出 n 个 'A' 的最少操作次数。

示例 1:

输入: 3

输出: 3

解释:

最初, 我们只有一个字符 'A'。

第 1 步, 我们使用 Copy All 操作。

第 2 步, 我们使用 Paste 操作来获得 'AA'。

第 3 步, 我们使用 Paste 操作来获得 'AAA'。

说明:

1. n 的取值范围是 [1, 1000] 。

思路

Intuition

We can break our moves into groups of (copy, paste, ..., paste). Let C denote copying and P denote pasting. Then for example, in the sequence of moves CPPCPPPPCP, the groups would be [CPP][CPPPP][CP].

Say these groups have lengths g_1, g_2, .... After parsing the first group, there are g_1 'A's. After parsing the second group, there are g_1 * g_2 'A's, and so on. At the end, there are g_1 * g_2 * ... * g_n 'A's.

We want exactly N = g_1 * g_2 * ... * g_n. If any of the g_i are composite, say g_i = p * q, then we can split this group into two groups (the first of which has one copy followed by p-1 pastes, while the second group having one copy and q-1 pastes).

Such a split never uses more moves: we use p+q moves when splitting, and pq moves previously. As p+q <= pq is equivalent to 1 <= (p-1)(q-1), which is true as long as p >= 2 and q >= 2.

Algorithm By the above argument, we can suppose g_1, g_2, ... is the prime factorization of N, and the answer is therefore the sum of these prime factors.

class Solution {
    public int minSteps(int n) {
        int ans = 0, d = 2;
        while (n > 1) {
            while (n % d == 0) {
                ans += d;
                n /= d;
            }
            d++;
        }
        return ans;
    }
}