650. 2 Keys Keyboard

最新推荐文章于 2025-08-24 15:22:07 发布

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最新推荐文章于 2025-08-24 15:22:07 发布

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文章标签： java c/c++

原文链接：http://www.cnblogs.com/ruruozhenhao/p/10517029.html

本文探讨了在初始状态只有一个字符'A'的记事本上，通过CopyAll和Paste操作达到指定数量'A'字符的最少步骤算法。提出了两种方法：动态规划(DP)和贪心算法，分析了各自的优劣及适用场景。

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Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step:

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

Given a number n. You have to get exactly n 'A' on the notepad by performing the minimum number of steps permitted. Output the minimum number of steps to get n 'A'.

Example 1:

Input: 3
Output: 3
Explanation:
Intitally, 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'.

Note:

The n will be in the range [1, 1000].

Approach #1: DP. [Java]

class Solution {
    public int minSteps(int n) {
        int[] dp = new int[n+1];
        
        for (int i = 2; i <= n; ++i) {
            dp[i] = i;
            for (int j = i-1; j > 1; --j) {
                if (i % j == 0) {
                    dp[i] = dp[j] + (i/j);
                    break;
                }
            }
        }
        
        return dp[n];
    }
}

Approach #2: Greedy. [C++]

    public int minSteps(int n) {
        int s = 0;
        for (int d = 2; d <= n; d++) {
            while (n % d == 0) {
                s += d;
                n /= d;
            }
        }
        return s;
    }

Analysis:

We look for a divisor d so that we can make d copies of (n / d) to get n. The process of making d copies takes d steps (1 step of copy All and d-1 steps of Paste)

We keep reducing the problem to a smaller one in a loop. The best cases occur when n is decreasing fast, and method is almost O(log(n)). For example, when n = 1024 then n will be divided by 2 for only 10 iterations, which is much faster than O(n) DP method.

The worst cases occur when n is some multiple of large prime, e.g. n = 997 but such cases are rare.

Reference:

https://leetcode.com/problems/2-keys-keyboard/discuss/105897/Loop-best-case-log(n)-no-DP-no-extra-space-no-recursion-with-explanation

https://leetcode.com/problems/2-keys-keyboard/discuss/105899/Java-DP-Solution

转载于:https://www.cnblogs.com/ruruozhenhao/p/10517029.html