Build Array Where You Can Find The Maximum Exactly K Comparisons

Problem

You are given three integers n, m and k. Consider the following algorithm to find the maximum element of an array of positive integers:

You should build the array arr which has the following properties:

arr has exactly n integers.
1 <= arr[i] <= m where (0 <= i < n).
After applying the mentioned algorithm to arr, the value search_cost is equal to k.

Return the number of ways to build the array arr under the mentioned conditions. As the answer may grow large, the answer must be computed modulo 10^9 + 7.

Example 1:

Input: n = 2, m = 3, k = 1
Output: 6
Explanation: The possible arrays are [1, 1], [2, 1], [2, 2], [3, 1], [3, 2] [3, 3]

Example 2:

Input: n = 5, m = 2, k = 3
Output: 0
Explanation: There are no possible arrays that satisify the mentioned conditions.

Example 3:

Input: n = 9, m = 1, k = 1
Output: 1
Explanation: The only possible array is [1, 1, 1, 1, 1, 1, 1, 1, 1]

Constraints:

1 <= n <= 50
1 <= m <= 100
0 <= k <= n

Solution

class Solution {
    private static final int MOD = 1_000_000_007;

    public int numOfArrays(int n, int m, int k) {
        long[][] ways = new long[m + 1][k + 1];
        long[][] sums = new long[m + 1][k + 1];
        for (int max = 1; max <= m; max++) {
            ways[max][1] = 1;
            sums[max][1] = ways[max][1] + sums[max - 1][1];
        }
        for (int count = 2; count <= n; count++) {
            long[][] ways2 = new long[m + 1][k + 1];
            long[][] sums2 = new long[m + 1][k + 1];
            for (int max = 1; max <= m; max++) {
                for (int cost = 1; cost <= k; cost++) {
                    long noCost = max * ways[max][cost] % MOD;
                    long newCost = sums[max - 1][cost - 1];
                    ways2[max][cost] = (noCost + newCost) % MOD;
                    sums2[max][cost] = (sums2[max - 1][cost] + ways2[max][cost]) % MOD;
                }
            }
            ways = ways2;
            sums = sums2;
        }
        return (int) sums[m][k];
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).