LeetCode Problem 3317

Difficulty: 🔴 Hard
Topics: Math, Dynamic Programming, Combinatorics

Solution

LeetCode 3317 - Find the Number of Possible Ways for an Event

Problem Understanding

The problem describes an event with n performers and x available stages. Each performer must be assigned to exactly one stage. Multiple performers may share the same stage, which means they form a band together. Some stages may remain unused.

After the performers are grouped into bands according to their assigned stages, each non-empty band receives a score from the jury. Every score is an integer in the range [1, y].

We must count the total number of distinct valid events.

Two events are considered different if either:

1. At least one performer is assigned to a different stage.
2. At least one band receives a different score.

The important detail is that stages are distinct. Assigning performers to stage 1 versus stage 2 creates different configurations even if the performer groups are structurally identical.

The constraints are:

• 1 <= n, x, y <= 1000

This immediately rules out brute-force enumeration. Even just assigning performers to stages has x^n possibilities, which becomes astronomically large when n = 1000.

The problem is fundamentally a combinatorics counting problem combined with dynamic programming.

The key observations are:

• We only care about non-empty stages because only non-empty stages become bands.

• If exactly k stages are used:

• We choose which k stages are active.

• We partition the n performers into k non-empty groups.

• We assign each group to one of the chosen stages.

• We assign a score to each band.

This naturally leads to Stirling numbers of the second kind.

Important edge cases include:

• y = 1, meaning all bands receive the same score.
• x > n, where many stages must remain empty.
• n = 1, the smallest possible performer count.
• Very large values near 1000, where efficiency and modular arithmetic become essential.

Approaches

Brute Force Approach

A brute-force solution would try every possible assignment of performers to stages.

Each performer has x choices, so there are:

$x^n$

possible assignments.

For every assignment, we would:

1. Determine which stages are non-empty.
2. Count the number of bands.
3. Enumerate all possible score assignments.

If an assignment uses k stages, there are:

$y^k$

ways to assign scores.

Although this method is conceptually straightforward, it is computationally impossible for n = 1000.

Key Insight

Instead of assigning performers individually, we count configurations based on the number of non-empty stages.

Suppose exactly k stages are used.

We need to count:

1. Ways to choose the k active stages.
2. Ways to partition n performers into k non-empty groups.
3. Ways to match groups to stages.
4. Ways to assign scores to bands.

The number of ways to partition n labeled performers into k non-empty unlabeled groups is the Stirling number of the second kind:

$S(n,k)$

After partitioning:

• We choose and order k stages from x stages:

$P(x,k)=\frac{x!}{(x-k)!}$

• Each band independently receives one of y scores:

$y^k$

So the total becomes:

$\sum_{k=1}^{\min(n,x)} S(n,k) \cdot P(x,k) \cdot y^k$

This transforms the problem into efficient combinatorics and DP.

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(x^n)	O(n)	Enumerates every performer assignment
Optimal	O(n * min(n, x))	O(n * min(n, x))	Uses Stirling numbers and combinatorics

Algorithm Walkthrough

Step 1: Define the DP for Stirling Numbers

We use the recurrence relation for Stirling numbers of the second kind:

$S(n,k)=S(n-1,k-1)+k\cdot S(n-1,k)$

Interpretation:

• The new performer either forms a new group.
• Or joins one of the existing k groups.

We build a DP table where:

• dp[i][j] = number of ways to partition i performers into j non-empty groups.

Step 2: Initialize Base Cases

The fundamental base case is:

$S(0,0)=1$

This represents one way to partition zero elements into zero groups.

All other impossible states remain zero.

Step 3: Fill the DP Table

For every i from 1 to n:

• For every j from 1 to i:

• Apply the recurrence relation.

All calculations are performed modulo:

$10^9+7$

Step 4: Compute Falling Factorials

We need:

$P(x,k)=x(x-1)(x-2)\cdots(x-k+1)$

Instead of computing factorials repeatedly, we incrementally maintain the current permutation count.

Step 5: Compute Powers of y

We also need:

$y^k$

We compute this iteratively while looping over k.

Step 6: Sum Contributions

For each valid k:

1. dp[n][k] gives partitions into k groups.
2. perm gives stage assignments.
3. pow_y gives score assignments.

Multiply them together and add to the answer.

Why it works

The algorithm exhaustively counts every valid event exactly once.

• S(n,k) partitions performers into k non-empty bands.
• P(x,k) assigns those bands to distinct labeled stages.
• y^k assigns scores independently to each band.

Every valid event corresponds to one unique combination of these three components, so the counting is complete and non-overlapping.

Python Solution

class Solution:
    def numberOfWays(self, n: int, x: int, y: int) -> int:
        MOD = 10**9 + 7

        # dp[i][j] = Stirling number of the second kind S(i, j)
        dp = [[0] * (n + 1) for _ in range(n + 1)]
        dp[0][0] = 1

        for i in range(1, n + 1):
            for j in range(1, i + 1):
                dp[i][j] = (
                    dp[i - 1][j - 1]
                    + j * dp[i - 1][j]
                ) % MOD

        answer = 0
        permutation = 1
        power_y = 1

        limit = min(n, x)

        for k in range(1, limit + 1):
            permutation = permutation * (x - k + 1) % MOD
            power_y = power_y * y % MOD

            ways = (
                dp[n][k]
                * permutation
                % MOD
                * power_y
            ) % MOD

            answer = (answer + ways) % MOD

        return answer

The implementation begins by constructing the Stirling number DP table.

The recurrence:

dp[i][j] = dp[i - 1][j - 1] + j * dp[i - 1][j]

matches the combinatorial interpretation directly.

After the DP table is built, the algorithm iterates over every possible number of non-empty stages k.

Instead of recomputing permutations and powers from scratch, the code maintains:

• permutation = P(x, k)
• power_y = y^k

incrementally.

For each k, the total contribution is:

dp[n][k] * permutation * power_y

which is added into the final answer modulo 10^9 + 7.

Go Solution

func numberOfWays(n int, x int, y int) int {
	const MOD int = 1_000_000_007

	dp := make([][]int, n+1)
	for i := range dp {
		dp[i] = make([]int, n+1)
	}

	dp[0][0] = 1

	for i := 1; i <= n; i++ {
		for j := 1; j <= i; j++ {
			dp[i][j] = (dp[i-1][j-1] + j*dp[i-1][j]) % MOD
		}
	}

	answer := 0
	permutation := 1
	powerY := 1

	limit := n
	if x < limit {
		limit = x
	}

	for k := 1; k <= limit; k++ {
		permutation = permutation * (x - k + 1) % MOD
		powerY = powerY * y % MOD

		ways := dp[n][k]
		ways = ways * permutation % MOD
		ways = ways * powerY % MOD

		answer = (answer + ways) % MOD
	}

	return answer
}

The Go implementation follows the same logic as the Python solution.

One important difference is integer handling. Since Go uses fixed-size integers, every multiplication is immediately reduced modulo MOD to prevent overflow growth.

The DP table is implemented using slices of slices, which naturally model a 2D array.

Worked Examples

Example 1

Input:

n = 1, x = 2, y = 3

We compute:

k	S(1,k)	P(2,k)	3^k	Contribution
1	1	2	3	6

Final answer:

6

Interpretation:

• Performer chooses stage 1 or stage 2.
• The resulting band gets score 1, 2, or 3.

Total:

2 * 3 = 6

Example 2

Input:

n = 5, x = 2, y = 1

Since every band must receive score 1, scoring contributes nothing extra.

We compute:

k	S(5,k)	P(2,k)	1^k	Contribution
1	1	2	1	2
2	15	2	1	30

Total:

2 + 30 = 32

Example 3

Input:

n = 3, x = 3, y = 4

We compute:

k	S(3,k)	P(3,k)	4^k	Contribution
1	1	3	4	12
2	3	6	16	288
3	1	6	64	384

Final answer:

12 + 288 + 384 = 684

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * min(n, x))	DP table construction dominates
Space	O(n^2)	Stores Stirling DP table

The DP table contains approximately n^2 / 2 meaningful states, and each state is computed in constant time.

The final summation loop runs only up to min(n, x).

Because n <= 1000, this complexity is efficient enough.

Test Cases

sol = Solution()

# Provided examples
assert sol.numberOfWays(1, 2, 3) == 6          # Single performer
assert sol.numberOfWays(5, 2, 1) == 32         # Only one possible score
assert sol.numberOfWays(3, 3, 4) == 684        # Multiple stages and scores

# Minimum boundary
assert sol.numberOfWays(1, 1, 1) == 1          # One performer, one stage, one score

# More stages than performers
assert sol.numberOfWays(2, 5, 1) == 25         # Equivalent to 5^2 assignments

# Single stage only
assert sol.numberOfWays(4, 1, 7) == 7          # Everyone forced onto same stage

# Single score only
assert sol.numberOfWays(3, 2, 1) == 8          # Pure stage assignments

# Distinct performers with exact stage usage
assert sol.numberOfWays(2, 2, 2) == 12

# Large y value
assert sol.numberOfWays(1, 3, 1000) == 3000    # One band, many score choices

# Stress-style sanity check
result = sol.numberOfWays(1000, 1000, 1000)
assert isinstance(result, int)                 # Ensures large input runs successfully

Test Summary

Test	Why
(1,2,3)	Verifies smallest non-trivial case
(5,2,1)	Verifies handling when score choices disappear
(3,3,4)	Validates full combinatorial behavior
(1,1,1)	Smallest legal input
(2,5,1)	Ensures unused stages are handled correctly
(4,1,7)	Verifies all performers forced onto one stage
(3,2,1)	Checks pure assignment counting
(2,2,2)	Small balanced case
(1,3,1000)	Verifies large scoring range
(1000,1000,1000)	Stress test for performance

Edge Cases

Case 1: More stages than performers

Consider:

n = 2, x = 100

At most two stages can be non-empty because there are only two performers.

A buggy implementation might incorrectly iterate beyond feasible group counts. The solution avoids this by looping only up to:

min(n, x)

which guarantees valid states only.

Case 2: Only one stage

Consider:

n = 1000, x = 1

Every performer must go onto the same stage.

There is exactly one possible grouping structure, but the band can still receive any score from 1 to y.

The algorithm correctly computes:

S(n,1) * P(1,1) * y

which simplifies to:

1 * 1 * y

Case 3: Only one possible score

Consider:

y = 1

In this case, score assignment contributes no additional multiplicity.

A buggy implementation might still attempt complicated score enumeration. Our formula naturally reduces because:

$1^k=1$

for all k.

The implementation handles this automatically without any special-case logic.