> For the complete documentation index, see [llms.txt](https://protegejj.gitbook.io/algorithm-practice/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://protegejj.gitbook.io/algorithm-practice/leetcode/dynamic-programming/650-2-keys-keyboard.md).

# 650     2 Keys Keyboard

## 650. [2 Keys Keyboard](https://leetcode.com/problems/2-keys-keyboard/description/)

## 1. Question

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].

## **2. Implementation**

**(1) DP**

```java
class Solution {
    public int minSteps(int n) {
        if (n <= 1) {
            return 0;
        }
        int[] dp = new int[n + 1];

        for (int i = 1; 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];
    }
}
```

## 3. Time & Space Complexity

**DFS + Memoization**: 时间复杂度O(k ^(mn)), k是offer的数量, m是item的个数，n是每个item要买的个数，空间复杂度O(mn)?