<p>There is only one character <code>&#39;A&#39;</code> on the screen of a notepad. You can perform one of two operations on this notepad for each step:</p>

<ul>
	<li>Copy All: You can copy all the characters present on the screen (a partial copy is not allowed).</li>
	<li>Paste: You can paste the characters which are copied last time.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the minimum number of operations to get the character</em> <code>&#39;A&#39;</code> <em>exactly</em> <code>n</code> <em>times on the screen</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> Initially, we have one character &#39;A&#39;.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get &#39;AA&#39;.
In step 3, we use Paste operation to get &#39;AAA&#39;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

class Solution {
public:
    long long t[2002][2002][2];
    Solution(){
        memset(t , -1 , sizeof(t));
    }
    long long solve(int i , int n , int prev , int copied){
        if(i == n){
            return 0;
        }
        if(i > n){
            return INT_MAX-1;
        }
        if(t[i][copied][prev] != -1){
            return t[i][copied][prev];
        }
        if(prev == 0){
            return t[i][copied][prev] = 2 + solve(i+1 , n , 1 , 1);
        }
        else{
            return t[i][copied][prev] = min(1 + solve(i+copied , n , 1 , copied) , 2 + solve(i+i , n , 1 , i));
        }
    }

    int minSteps(int n) {
        return solve(1 , n , 0 , 0);
    }
};
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