<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 {
private:
    int solve(int i, int len, int n, vector<vector<int>> &dp){
        if(i==n) return 0;
        if(i>n) return INT_MAX;
        if(dp[i][len] != -1) return dp[i][len];
        int paste = solve(i+len, len, n, dp);
        if(paste < INT_MAX) paste++;
        int copy = INT_MAX;
        if((n-i) >= i) {
            copy = solve(i+i, i, n, dp);
            if(copy < INT_MAX-2) copy+=2;
        }
        return dp[i][len] = min(copy, paste);
    }
public:
    int minSteps(int n) {
        if(n==1) return 0;
        int cnt=0;
        for(int i=1; i<=n; i++){
            if(n%i == 0) cnt++;
        }
        if(cnt==2) return n;
        vector<vector<int>> dp(n, vector<int>(n, -1));
        return 1+solve(1, 1, n, dp);
    }
};

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