LeetCode 650. 2 Keys Keyboard

原题链接在这里：https://leetcode.com/problems/2-keys-keyboard/description/

题目：

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

1. Copy All: You can copy all the characters present on the notepad (partial copy is not allowed).
2. 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:

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

题解：

Set some small examples from 1 or 2 A. Then found out this could be sloved by DP.

Let dp[i] denotes up to i A, the minimum number of steps needed. 

初始化 i 从2到n赋值 i 本身，代表每次直接paste 一个'A'. i = 1时, dp[i] = 0, 本身就有一个'A'.

状态转移，对于每一个j 在(1,n)区间内 如果i%j == 0 并且dp[j] + i/j比 dp[i]小就更新dp[i].

表示重新 Copy All, 用1次操作, 再paste (i/j-1)次. 公用i/j-1+1 = i/j次操作. j不用减到1因为最开始的dp[i]就是通过1个'A'算出来的.

答案dp[n]. 

Time Complexity: O(n^2).

Space: O(n).

AC 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] = Math.min(dp[i], dp[j]+i/j);
                }
            }
        }
        return dp[n];
    }
}

类似4 Keys Keyboard.