<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 minSteps(int n) {
    //    dp
    vector<vector<int>> dp(n, vector<int>(n,1000));
    int i,j;
    for(i = 0; i < n; i++){
        dp[i][0] = i;
        dp[0][i] = 1000;

    }
    int mini;
    dp[0][0]=1;
    for(i = 1; i < n; ++i){ 
        mini = 1000;
        for(j = 0; j < i; j++){ // should only cover bot left triangle
           dp[i][j] = dp[i-j-1][j]+1; 
           mini = min(dp[i][j], mini);
        }
        // assume copy of new number
        dp[i][i] = mini+1;
    }
    // for( i = 0; i < n; i++){
    //     // for(j = 0; j < n; j++){
    //         // cout << dp[i][j] << " ";
    //     }
    //     // cout << endl;
    // }
    return dp[n-1][n-1]-1;


    }
};
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