<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 {
 public:
  int numOfArrays(int n, int m, int k) {
    constexpr int kMod = 1'000'000'007;
    // dp[i][j][k] := the number of ways to build an array of length i, where j
    // is the maximum number and k is `search_cost`
    vector<vector<vector<int>>> dp(
        n + 1, vector<vector<int>>(m + 1, vector<int>(k + 1)));

    for (int j = 1; j <= m; ++j)
      dp[1][j][1] = 1;

    for (int i = 2; i <= n; ++i)                 // for each length
      for (int j = 1; j <= m; ++j)               // for each max value
        for (int cost = 1; cost <= k; ++cost) {  // for each cost
          // 1. Appending any of [1, j] in the i-th position doesn't change the
          //    maximum and cost.
          dp[i][j][cost] = static_cast<long>(j) * dp[i - 1][j][cost] % kMod;
          // 2. Appending j in the i-th position makes j the new max and cost 1.
          for (int prevMax = 1; prevMax < j; ++prevMax) {
            dp[i][j][cost] += dp[i - 1][prevMax][cost - 1];
            dp[i][j][cost] %= kMod;
          }
        }

    int ans = 0;
    for (int j = 1; j <= m; ++j) {
      ans += dp[n][j][k];
      ans %= kMod;
    }
    return ans;
  }
};
13500