<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 mod = 1000000007;

    long long modExp(long long base, int exp) {
        long long result = 1;
        while (exp > 0) {
            if (exp % 2 == 1) {
                result = (result * base) % mod;
            }
            base = (base * base) % mod;
            exp /= 2;
        }
        return result;
    }

    long long func(int i,int ma, int n, int m, int k, vector<vector<vector<long long>>>& dp) {
        // Base case: No elements left to place
        if (n==i) {
            return (ma <= m && k == 0) ? 1 : 0;
        }

        // Check memoization table
        if (dp[i][ma][k] != -1) return dp[i][ma][k];

        long long ans=0;
        for (int j=1;j<=ma;j++){
            ans+=func(i+1,ma,n,m,k,dp)%mod;
        }
        if (k>0){
        for (int j=ma+1;j<=m;j++){
            ans+=func(i+1,j,n,m,k-1,dp)%mod;
        }
        }
        // Memoize and return the result
        return dp[i][ma][k] = ans;
    }

    int numOfArrays(int n, int m, int k) {
        int ans = 0;
        vector<vector<vector<long long>>> dp(n + 1, vector<vector<long long>>(m + 1, vector<long long>(k + 1, -1)));
        for (int i = 1; i <= m; ++i) {
            ans = (ans + func(1,i, n , m, k - 1, dp)) % mod;
        }
        return ans;
    }
};

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