<p>There are <code>n</code> uniquely-sized sticks whose lengths are integers from <code>1</code> to <code>n</code>. You want to arrange the sticks such that <strong>exactly</strong> <code>k</code>&nbsp;sticks are <strong>visible</strong> from the left. A stick&nbsp;is <strong>visible</strong> from the left if there are no <strong>longer</strong>&nbsp;sticks to the <strong>left</strong> of it.</p>

<ul>
	<li>For example, if the sticks are arranged <code>[<u>1</u>,<u>3</u>,2,<u>5</u>,4]</code>, then the sticks with lengths <code>1</code>, <code>3</code>, and <code>5</code> are visible from the left.</li>
</ul>

<p>Given <code>n</code> and <code>k</code>, return <em>the <strong>number</strong> of such arrangements</em>. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, k = 2
<strong>Output:</strong> 3
<strong>Explanation:</strong> [<u>1</u>,<u>3</u>,2], [<u>2</u>,<u>3</u>,1], and [<u>2</u>,1,<u>3</u>] are the only arrangements such that exactly 2 sticks are visible.
The visible sticks are underlined.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 5, k = 5
<strong>Output:</strong> 1
<strong>Explanation:</strong> [<u>1</u>,<u>2</u>,<u>3</u>,<u>4</u>,<u>5</u>] is the only arrangement such that all 5 sticks are visible.
The visible sticks are underlined.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 20, k = 11
<strong>Output:</strong> 647427950
<strong>Explanation:</strong> There are 647427950 (mod 10<sup>9 </sup>+ 7) ways to rearrange the sticks such that exactly 11 sticks are visible.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>1 &lt;= k &lt;= n</code></li>
</ul>

class Solution:
    def rearrangeSticks(self, n: int, k: int) -> int:
        # dp[n_visible] = # of ways to place remaining n_remain sticks 
        # if we need to make n_visible of them visible from left
        # how many sticks remained depends on which iteration we are currently at
        dp = [0] * (k + 1)
        
        # basecase: when there is no stick remain -> 1 if no more sticks need to be visible
        dp[0] = 1

        # tabulate
        for n_remain in range(1, n + 1):
            # since n_remain > 0, so when n_visible = 0 -> no way to place correct stick
            new_dp = [0] * (k + 1)
            
            # n_visible must be <= n_remain or k, if overbound set to 0 (done by line 14)
            for n_visible in range(1, min(n_remain, k) + 1):
                # place max sticks at the end of remaining sticks
                new_dp[n_visible] = dp[n_visible - 1]

                # place non-max sticks at the end of remaining sticks
                new_dp[n_visible] += (n_remain - 1) * dp[n_visible]

                new_dp[n_visible] %= (10 ** 9 + 7)

            dp = new_dp

        return dp[-1]

        

        
       
            


            
            



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