<p>You are given three integers <code>n</code>, <code>m</code> and <code>k</code>. Consider the following algorithm to find the maximum element of an array of positive integers:</p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/02/e.png" style="width: 424px; height: 372px;" />
<p>You should build the array arr which has the following properties:</p>

<ul>
	<li><code>arr</code> has exactly <code>n</code> integers.</li>
	<li><code>1 &lt;= arr[i] &lt;= m</code> where <code>(0 &lt;= i &lt; n)</code>.</li>
	<li>After applying the mentioned algorithm to <code>arr</code>, the value <code>search_cost</code> is equal to <code>k</code>.</li>
</ul>

<p>Return <em>the number of ways</em> to build the array <code>arr</code> under the mentioned conditions. As the answer may grow large, the answer <strong>must be</strong> computed modulo <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> n = 2, m = 3, k = 1
<strong>Output:</strong> 6
<strong>Explanation:</strong> The possible arrays are [1, 1], [2, 1], [2, 2], [3, 1], [3, 2] [3, 3]
</pre>

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

<pre>
<strong>Input:</strong> n = 5, m = 2, k = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no possible arrays that satisfy the mentioned conditions.
</pre>

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

<pre>
<strong>Input:</strong> n = 9, m = 1, k = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only possible array is [1, 1, 1, 1, 1, 1, 1, 1, 1]
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>1 &lt;= m &lt;= 100</code></li>
	<li><code>0 &lt;= k &lt;= n</code></li>
</ul>

from functools import cache

class Solution:
    MOD = 10 ** 9 + 7

    def numOfArrays(self, n: int, m: int, k: int) -> int:
        if k > n or k > m:
            return 0

        return Solution.recurseOpen(n, m, k)


    @cache
    def recurseOpen(n, m, k):
        """m may or may not be present in the array"""
        if k == n:
            return Solution.nCr(m, k)
        elif k == 0:
            return 0

        total = 0
        for maximum in range(k, m + 1):
            total += Solution.recurseExact(n, maximum, k)

        return total % Solution.MOD


    @cache
    def recurseExact(n, m, k):
        """m must be present in the array"""
        total = 0
        for firstMax in range(k, n + 1):
            continuations = m ** (n - firstMax)
            total += continuations * Solution.recurseOpen(firstMax - 1, m - 1, k - 1)

        return total % Solution.MOD


    @cache
    def nCr(n, r):
        if 2 * r > n:
            r = n - r

        total = 1
        for q in range(1, r + 1):
            total *= n
            total //= q
            n -= 1

        return total % Solution.MOD
32000