<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:
    int dp[1500][1500][5];
    int rec(int n,int ch =1, int cur =0, bool c = 0){
        if(n < ch)
        return 1e9;
        else if(n == ch)
        return 0;

        int &ret = dp[ch][cur][c];
        if(~ret)
        return ret;

        int ans = 1e9;
        if(c == 0)
        ans = min(ans,1+rec(n, ch,ch, 1 ));
        if(cur > 0)
        ans = min(ans, 1 + rec(n, ch+cur, cur, 0));
        return ret = ans;
    }
    
    int minSteps(int n) {
        memset(dp, -1, sizeof(dp));
        return rec(n);
    }
};
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