<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>

class Solution:
    def numOfArrays(self, n: int, m: int, k: int) -> int:
        MOD = 10**9 + 7
        
        # dp[i][p][v]: number of ways to form arrays of length i, search cost p, max v
        dp = [[[0] * (m + 1) for _ in range(k + 1)] for _ in range(n + 1)]
        
        # Base cases:
        for v in range(1, m + 1):
            dp[1][1][v] = 1  # One way to form an array of length 1 with cost 1 with max v
        
        # Fill DP table
        for i in range(2, n + 1):  # Length of the array
            for p in range(1, k + 1):  # Search cost
                for v in range(1, m + 1):  # Max element
                    # Case 1: Last element is not the max
                    dp[i][p][v] = dp[i-1][p][v] * v % MOD
                    
                    # Case 2: Last element is the new max
                    sum_prev_max = 0
                    for prev_max in range(1, v):
                        sum_prev_max = (sum_prev_max + dp[i-1][p-1][prev_max]) % MOD
                    dp[i][p][v] = (dp[i][p][v] + sum_prev_max) % MOD
        
        # Answer is the sum of dp[n][k][v] for all possible max values `v`
        result = 0
        for v in range(1, m + 1):
            result = (result + dp[n][k][v]) % MOD
            
        return result
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