<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) {
        const int MAX_CLIPBOARD_LEN = n >> 1;
        // dp[i][len of str in clipboard] = min ops
        vector<vector<int>> dp(n, vector<int>(MAX_CLIPBOARD_LEN + 1, 1000));
        dp[0][0] = 0;
        queue<pair<int, int>> q;
        q.push({0, 0});
        while (!q.empty()) {
            int i = q.front().first;
            int clipboardLen = q.front().second;
            q.pop();
            if (i == n - 1) {
                return dp[i][clipboardLen];
            }
            int ops = dp[i][clipboardLen] + 1;
            // copy
            int newClipboardLen = i + 1;
            if (newClipboardLen <= MAX_CLIPBOARD_LEN &&
                ops < dp[i][newClipboardLen]) {
                dp[i][newClipboardLen] = ops;
                q.push({i, newClipboardLen});
            }
            // paste
            int newI = i + clipboardLen;
            if (newI < n && ops < dp[newI][clipboardLen]) {
                dp[newI][clipboardLen] = ops;
                q.push({newI, clipboardLen});
            }
        }
        return 1000;
    }
};
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