<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 {
    /*
        From backtrack to dp
        State variables:
            1) i: current copied string length
            2) j: current string length on screen
            3) dp[i][j] is the minimum steps at (i,j)
        Recurrence relation:
            dp[i][j] = min(
                1 + dp[i][i+j], // paste only
                2 + dp[j][j+j]  // copy + paste
            )
        Base case:
            initial condition: dp[1][1] = 1 // copy the initial one 'A'
    */
    // bool debug = true;
    // inline static int casenum = 1;
public:
    int dp(vector<vector<int>>& memo, int n, int i, int j) {
        // i is the previously-copied length; j is the current string length
        if (j > n) return 1001;
        else if (j == n) return 0;
        // if (debug) cout << "i: " << i << " j: " << j << endl;
        if (memo[i][j]) return memo[i][j];
        memo[i][j] = min(
            1 + dp(memo, n, i, j+i), // paste only
            2 + dp(memo, n, j, j+j)  // copy current string and paste
        );
        return memo[i][j];
    }

    int minSteps(int n) {
        // if (debug) cout << "case " << casenum++ << endl;
        if (n == 1) return 0;
        // string length at most to 'n'; and copied-length at most to **n/2**
        vector<vector<int>> memo(n/2+1, vector<int>(n+1));
        // memo[1][1] = 2; // initial condition to copy and paste first 'A'
        // return dp(memo, n, 0, 1);
        return 1 + dp(memo, n, 1, 1);
    }
};
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