3317. Find the Number of Possible Ways for an Event

HardMathDynamic ProgrammingCombinatorics

Problem Description

You are given three integers n, x, and y.

An event is being held for n performers. When a performer arrives, they are assigned to one of the x stages. All performers assigned to the same stage will perform together as a band, though some stages might remain empty.

After all performances are completed, the jury will award each band a score in the range [1, y].

Return the total number of possible ways the event can take place.

Since the answer may be very large, return it modulo 10^9 + 7.

Note that two events are considered to have been held differently if either of the following conditions is satisfied:

Any performer is assigned a different stage.
Any band is awarded a different score.

Intuition

To solve this problem, we employ a dynamic programming approach. We need to calculate the number of ways to distribute n performers across x stages and then award scores to the bands. This can be thought of as a two-step process:

Distributing Performers: We use a DP table f[i][j] where f[i][j] represents the number of ways to arrange the first i performers into j stages. The base case is f[0][0] = 1 (one way to arrange zero performers), and we initialize other cases with zeros.

For distributing performers:
- If we assign the i-th performer to an already existing band: We have j choices (existing bands), so this contributes f[i - 1][j] * j.
- If we assign the i-th performer to a new band: We have (x - (j - 1)) choices (new stages), which adds f[i - 1][j - 1] * (x - (j - 1)).
The recurrence relation becomes:
```
f[i][j] = f[i - 1][j] * j + f[i - 1][j - 1] * (x - (j - 1))
```
Awarding Scores: Once we have distributed all performers, we can award scores. For each band j, there are y^j possible ways to assign scores. Hence, the total number of ways is summed up as:
```
Σ (f[n][j] * y^j)
```
Finally, as results can be large, we take results modulo 10^9 + 7.

This approach efficiently calculates the required number of ways to conduct the event while considering the constraints of the problem.

Learn more about Math, Dynamic Programming and Combinatorics patterns.

Solution Approach

The solution employs a dynamic programming approach to efficiently calculate the number of possible ways the event can take place. Here's a breakdown of the implementation:

Initialization:
- We define a 2D list f where f[i][j] represents the number of ways to assign the first i performers into j stages.
- Initialize f[0][0] = 1 because there is exactly one way to arrange zero performers into zero stages: do nothing. All other entries in the DP table f are initialized to zero.
Filling the DP Table:
- For each performer i from 1 to n (total performers), we iterate over each possible stage count j from 1 to x (total stages).
- For each i and j, update f[i][j] based on the choices:
  - If the i-th performer joins an existing stage, there are j choices, which contributes f[i-1][j] * j.
  - If the i-th performer initiates a new stage, there are (x - (j - 1)) choices, adding f[i-1][j-1] * (x - (j - 1)).
- The recurrence relation used is:
```
f[i][j] = f[i-1][j] * j + f[i-1][j-1] * (x - (j - 1))
```
- Each calculation is taken modulo 10^9 + 7 to avoid overflow.
Calculate Total Ways:
- Initialize ans to zero, representing the accumulated number of ways.
- For each number of stages j, compute the number of ways to award scores, which is y^j. Update ans by adding f[n][j] * y^j.
- Use a variable p to maintain the power y^j, updated in each iteration as p = (p * y) % mod.
- Sum up the results for all possible ways and take modulo 10^9 + 7.
Return the Result:
- The value stored in ans gives us the total number of ways to hold the event according to the conditions specified.

This dynamic programming approach ensures that we efficiently count all possible configurations of stage assignments and scoring.

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Example Walkthrough

Consider an example with n = 2, x = 2, and y = 2. This means there are 2 performers, 2 stages, and scores can range from 1 to 2.

Initialization: We create a DP table f where f[i][j] represents the number of ways to arrange i performers into j stages. Start with f[0][0] = 1.
Filling the DP Table: The table is filled as follows:
- For i = 1:
  - Stage count j = 1:
    f[1][1] = f[0][1] * 1 + f[0][0] * (2 - 0) = 0 * 1 + 1 * 2 = 2
  - Stage count j = 2:
    f[1][2] = f[0][2] * 2 + f[0][1] * (2 - 1) = 0 * 2 + 0 * 1 = 0
- For i = 2:
  - Stage count j = 1:
    f[2][1] = f[1][1] * 1 + f[1][0] * (2 - 0) = 2 * 1 + 0 * 2 = 2
  - Stage count j = 2:
    f[2][2] = f[1][2] * 2 + f[1][1] * (2 - 1) = 0 * 2 + 2 * 1 = 2
Calculate Total Ways:
- Initialize ans = 0 and set p = 1 for y^0.
- For j = 1:
  ans += f[2][1] * p = 2 * 1 = 2
  Update p = p * y % mod = 1 * 2 % (10^9 + 7) = 2
- For j = 2:
  ans += f[2][2] * p = 2 * 2 = 4
  Now ans = 2 + 4 = 6

Thus, there are 6 possible ways to assign performers and scores, modulo 10^9 + 7.

Solution Implementation

1class Solution:
2    def numberOfWays(self, n: int, x: int, y: int) -> int:
3        # Define the modulus value to ensure the results are within integer range
4        mod = 10**9 + 7
5      
6        # Initialize a 2D list (dynamic programming table) f, with dimensions (n+1) x (x+1),
7        # filled with zeros
8        f = [[0] * (x + 1) for _ in range(n + 1)]
9      
10        # Base case: there is one way to select 0 elements, which results in a sum of 0
11        f[0][0] = 1
12      
13        # Iterate over the number of items to consider for selection (i is the item index)
14        for i in range(1, n + 1):
15            # Iterate over the number of items that have been selected (j items selected)
16            for j in range(1, x + 1):
17                # Calculate the number of ways to form a subset of j items ending at i
18                f[i][j] = (f[i - 1][j] * j + f[i - 1][j - 1] * (x - (j - 1))) % mod
19      
20        # Initialize variables to calculate the final answer
21        ans, power = 0, 1
22      
23        # Iterate over each selected number to accumulate the answer with calculated powers of y
24        for j in range(1, x + 1):
25            # Update power by multiplying it with y and applying the modulus
26            power = power * y % mod
27            # Accumulate the result by considering the current subset count in f[n][j]
28            ans = (ans + f[n][j] * power) % mod
29      
30        # Return the calculated number of ways
31        return ans
32

1class Solution {
2    public int numberOfWays(int n, int x, int y) {
3        final int MOD = (int) 1e9 + 7; // Define modulus for operations to prevent overflow
4        long[][] dp = new long[n + 1][x + 1]; // DP array to store intermediate results
5        dp[0][0] = 1; // Base case initialization: 1 way to achieve 0 using 0 items
6      
7        // Calculate number of ways to achieve each count from 1 to n using up to x items
8        for (int i = 1; i <= n; ++i) {
9            for (int j = 1; j <= x; ++j) {
10                // Transition state: 
11                // Combine choosing or not choosing the current item (states from previous row)
12                dp[i][j] = (dp[i - 1][j] * j % MOD 
13                          + dp[i - 1][j - 1] * (x - (j - 1) % MOD)) % MOD;
14            }
15        }
16      
17        long result = 0, power = 1; // Result to store final answer, Power to handle y's multiplier
18      
19        // Calculate the final result combining all possible states with their respective power
20        for (int j = 1; j <= x; ++j) {
21            power = power * y % MOD; // Update power of y
22            result = (result + dp[n][j] * power) % MOD; // Accumulate result
23        }
24      
25        return (int) result; // Return the result as integer
26    }
27}
28

1#include <vector>
2#include <cstring> // for memset
3
4class Solution {
5public:
6    int numberOfWays(int n, int x, int y) {
7        const int MOD = 1e9 + 7;
8        // Declare a 2D array f with dimensions (n + 1) x (x + 1) to store the number of ways
9        std::vector<std::vector<long long>> f(n + 1, std::vector<long long>(x + 1, 0));
10      
11        // Base case: There's exactly one way to arrange 0 objects with 0 choices
12        f[0][0] = 1;
13
14        // Fill the table using dynamic programming
15        for (int i = 1; i <= n; ++i) {
16            for (int j = 1; j <= x; ++j) {
17                // Calculate the number of ways by considering two scenarios:
18                // 1. Choosing j from (i-1) objects
19                // 2. Choosing (j-1) from (i-1) objects
20                f[i][j] = (f[i - 1][j] * j % MOD + f[i - 1][j - 1] * (x - (j - 1) % MOD)) % MOD;
21            }
22        }
23
24        // Calculate the final answer using the computed values
25        long long answer = 0;
26        long long power = 1;
27
28        // Accumulate the total number of ways by considering contributions from all possible choices
29        for (int j = 1; j <= x; ++j) {
30            power = power * y % MOD; // Update the power of y
31            answer = (answer + f[n][j] * power) % MOD; // Add to the answer the weighted count
32        }
33
34        return answer;
35    }
36};
37

1function numberOfWays(n: number, x: number, y: number): number {
2    const mod = BigInt(10 ** 9 + 7);
3
4    // Initialize a 2D array `f` with dimensions (n+1) x (x+1) filled with BigInt '0'.
5    // This array stores computed values for dynamic programming.
6    const f: bigint[][] = Array.from({ length: n + 1 }, () => Array(x + 1).fill(0n));
7    f[0][0] = 1n; // Base case: there's one way to choose zero elements
8
9    // Populate the dynamic programming table.
10    for (let i = 1; i <= n; ++i) {
11        for (let j = 1; j <= x; ++j) {
12            // Calculate the number of ways to choose `j` elements from `i` items.
13            // Combination logic with permutations and previous state values.
14            f[i][j] = (f[i - 1][j] * BigInt(j) + f[i - 1][j - 1] * BigInt(x - (j - 1))) % mod;
15        }
16    }
17
18    let ans = 0n; // Initialize the result as BigInt zero
19    let p = 1n; // Initialize power multiplier for `y`
20
21    // Calculate the desired result using the computed values from the `f` array.
22    for (let j = 1; j <= x; ++j) {
23        p = (p * BigInt(y)) % mod; // Compute y^j modulo `mod`
24        // Sum up the number of ways multiplied by the power of `y`.
25        ans = (ans + f[n][j] * p) % mod;
26    }
27
28    return Number(ans); // Convert the result to a regular number before returning
29}
30

Time and Space Complexity

The time complexity of the code is O(n * x) because it involves two nested loops: the outer loop iterates over n and the inner loop over x. Each iteration performs a constant amount of work.

The space complexity is also O(n * x) since it requires a 2D list f with dimensions (n+1) * (x+1) to store intermediate results.

Learn more about how to find time and space complexity quickly.

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