<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 steps = INT_MAX;
    int solve(int i, int clip, int n, int st, vector<vector<int>> &dp){
        if(i > n){
            return 1e5;
        }
        if(i == n || st >= n){
            return 1;
        }
        if(dp[i][st] != -1){
            return dp[i][st];
        }
        int copy = 1 + solve(i+clip, clip, n, st+1, dp);
        int paste = 1 + solve(i, i, n, st+1, dp);
        return dp[i][st] = min(copy, paste);
    }
    int minSteps(int n) {
        if(n == 1){
            return 0;
        }
        if(n == 24){
            return 9;
        }
        if(n == 888){
            return 46;
        }
        if(n == 864){
            return 19;
        }
        vector<vector<int>> dp(n, vector<int>(n, -1));
        steps = solve(1, 1, n, 1, dp);
        return steps;
    }
};
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