<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 {
  vector<vector<int>> dp_;  // [curr_cnt][prev_cpy]
  int n_;

  void init(int n) {
    dp_.resize(n, vector<int>(n / 2 + 1, -1));
    n_ = n;
  }

  int dp(int curr_cnt, int prev_cpy) {
    if (curr_cnt > n_) {
      return INT_MAX;
    }
    if (curr_cnt == n_) {
      return 0;
    }
    int ans = dp_.at(curr_cnt).at(prev_cpy);
    if (ans == -1) {
      ans = INT_MAX;
      for (int curr_cpy : {
          prev_cpy/* either reuse prev_cpy */,
          curr_cnt/* or make a new copy-all before pasting */}) {
        int next_sol = dp(curr_cnt + curr_cpy, curr_cpy);
        if (next_sol != INT_MAX) {
          ans = min(ans, 1 + next_sol + (curr_cpy == curr_cnt));
        }
      }
      assert(0 < ans);
      dp_.at(curr_cnt).at(prev_cpy) = ans;
    }
    return ans;
  }

public:
  int minSteps(int n) {
    if (n == 1) {
      return 0;
    }
    init(n);
    return dp(1, 1);
  }
};
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