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We can use dynamic programming to solve this problem by maintaining a 3D array where dp[i][j][k] represents the number of ways to construct an array of length i, with values up to j and exactly k comparisons.
Recurrence relations can be defined by considering whether you want to extend the array by a value that does or doesn't increase the number of comparisons. Use modulo operations to manage large numbers.
Time Complexity: O(n * m^2 * k). Space Complexity: O(n * m * k).
1#include <vector>
2using namespace std;
3
4const int MOD = 1000000007;
5
6int numOfArrays(int n, int m, int k) {
7    vector<vector<vector<int>>> dp(n+1, vector<vector<int>>(m+1, vector<int>(k+1, 0)));
8    for (int j = 1; j <= m; ++j) dp[1][j][1] = 1;
9    for (int i = 2; i <= n; ++i) {
10        for (int j = 1; j <= m; ++j) {
11            for (int z = 1; z <= k; ++z) {
12                dp[i][j][z] = ((long long)j * dp[i-1][j][z]) % MOD;
13                for (int p = 1; p < j; ++p) {
14                    dp[i][j][z] = (dp[i][j][z] + dp[i-1][p][z-1]) % MOD;
15                }
16            }
17        }
18    }
19    int totalWays = 0;
20    for (int j = 1; j <= m; ++j) {
21        totalWays = (totalWays + dp[n][j][k]) % MOD;
22    }
23    return totalWays;
24}
25Following a similar pattern, the C++ solution uses nested loops to iterate over the dimensions of the problem. The outer loop iterates over the number of elements, the second outer loop iterates over the possible maximum value, and the innermost loop iterates over the number of comparisons. At each step, it calculates solutions based on whether the current number extends the max for the count of comparisons or fits in the current max formed so far.
This approach uses recursion with memoization to avoid recomputing the solutions for subproblems. The recursive function attempts to build the array incrementally by deciding, step by step, which value to add, and whether it will increase the maximum so far. By caching previously computed results, it reduces repeated calculations.
Time Complexity: O(n * m * k). Space Complexity: O(n * m * k) due to recursion stack and memoization storage.
1const MOD = 10 ** 9 +
The JavaScript solution utilizes a depth-first search assisted by a 3D memoization table to efficiently compute each state space. It simplifies the constraints by representing each possible array configuration analytically, making determinations at each recursion call to update necessary counts and conditions.