<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[i] -> number of times to get n A's on the screen 
        // if i is odd return i 
        // if i is even, either we can do i/2 + 1
        unordered_map<int, vector<int>> multiples; 
        for (int i = 2; i < n; i++) {
            int cur = i;
            cur += i; 
            while (cur <= 1000) {
                multiples[cur].push_back(i);
                cur += i; 
            }
        }
        
        vector<int> dp(n+1); 
        dp[1] = 0; 
        int min_val; 
        for (int i = 2; i <= n; i++) {
            // cout << i << endl; 
            min_val = i;
            for (auto& multiple : multiples[i]) {
                // cout << multiple << endl; 
                int num_times = (i - multiple) / multiple; 
                int cost = dp[multiple] + 1 + num_times;
                // cout << dp[multiple] << " " << cost << endl; 
                min_val = min(min_val, cost);
            }
            // cout << endl; 
            dp[i] = min_val; 
        }
        
        // for (auto& x : dp) {
        //     cout << x << " ";
        // }
        return dp[n];
        
    }
};
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