> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/dynamic-programming/2-keys-keyboard.md).

# 2 Keys Keyboard

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].

```java
// DP solution
class Solution {
    public int minSteps(int n) {
        int dp[] = new int[n + 1];
        dp[1] = 0;
        for (int i = 2; i <= n; i++) {
            dp[i] = i;// because you can just paste 'A' i times to get i*A
            for (int j = i/2; j >=1; j--) {
                if (i % j == 0){
                    dp[i] = Math.min(dp[i], dp[j] + i / j);
                    break;
                }
            }
        }
        return dp[n];
    }
}

/*
To get 81 we check
if (81 % 2 ==0) No
if (81 % 3 ==0) Yes
So we need to copy 81/3 = 27 'A's three times (3)
Now to get 27 'A's, we need to copy 27/3= 9 'A's three times (3)
To get 9 'A's, we need to copy 9/3=3 'A's three times (3)
And to get 3 'A's, we need to copy 3/3=1 'A's three times (3)
Final answer is 3+3+3+3 = 12
*/
class Solution {
    public int minSteps(int n) {
        int res = 0;
        for (int i = 2; i <= n; i++) {
            while (n % i == 0) {
                res += i;
                n = n / i;
            }
        }
        return res;
    }
}
```


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