650. 2 Keys Keyboard

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].

EXPLANATION:

1.f(1) = 0

f(2) = f(1)+1+1 一次为copy，一次为paste

f(3) = f(1)+1+1+1 一次为copy，一次为paste，paste

f（4）就有很多种了，可以一直paste就是4次，还有就是f(4) = f(2)+1+1因为f（2）加一次copy再加一次paste

f(n) = f(n/i)+n/i

SOLUTION:

``````class Solution {
public int minSteps(int n) {
if(n <= 1) {
return 0;
}

int result = n;
for(int i = n - 1; i > 1; i--) {
if(n % i == 0) {
result = Math.min(result, minSteps(n / i) + i);
}
}
return result;
}
}

import java.util.*;
class Solution {
public int minSteps(int n) {
return minStepsHelper(n);
}

public static int minStepsHelper(int n) {
if(n==1) return 0;
if(n==2) return 2;
if(n==3) return 3;
if(n==4) return 4;
if(n==5) return 5;
Iterator<Map.Entry<Integer, Integer>> iterator = helper(n).entrySet().iterator();
int result = Integer.MAX_VALUE;
while (iterator.hasNext()){
Map.Entry<Integer, Integer> next = iterator.next();
int a =Integer.MAX_VALUE;
int b =Integer.MAX_VALUE;
if(next.getKey()!=1){
a = minStepsHelper(next.getKey())+next.getValue();
b = minStepsHelper(next.getValue())+next.getKey();
result = Math.min(result,a>b?b:a);
}else {
result = Math.min(result,minStepsHelper(next.getKey())+next.getValue());
}
}
return result;
}
public static Hashtable<Integer,Integer> helper(int n){
Hashtable<Integer,Integer> result = new Hashtable<>();
for(int i = 1;i<=Math.sqrt(n);i++){
if(n%i==0) result.put(i,n/i);
}
return result;
}

}
``````