<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: #iterative, O(k) space
    def rearrangeSticks(self, n: int, k: int) -> int:
        prev=[0]*(k+1)
        prev[1]=1#base case when n=1
        curr=prev.copy()
        mod=10**9+7
        for i in range(2,n+1):
            for x in range(1,k+1):
                curr[x]=((i-1)*prev[x]+prev[x-1])%mod
            prev=curr.copy()
        return curr[-1]
class Solution: #iterative
    def rearrangeSticks(self, n: int, k: int) -> int:
        dp={(1,1):1} #base case
        mod=10**9+7
        for i in range(2,n+1):
            for x in range(1,k+1):
                dp[(i,x)]=((i-1)*dp.get((i-1,x),0)+dp.get((i-1,x-1),0))%mod
        return dp[(n,k)]
# class Solution: #recursive solution->doesn't get accepted bcz of recursion depth 
#     def rearrangeSticks(self, n: int, k: int) -> int:
#         dp={}
#         mod=10**9+7
#         def dfs(i,k):
#             if i==k:return 1
#             if i==0 or k==0 or i<k:return 0
#             if (i,k) in dp:return dp[(i,k)]
#             dp[(i,k)]=(i-1)*dfs(i-1,k)+dfs(i-1,k-1)
#             return dp[(i,k)]%mod
#         return dfs(n,k)
3482