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Problem solution · Python

Find the Number of Possible Ways for an Event

Find the Number of Possible Ways for an Event: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
37 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Find the Number of Possible Ways for an Event, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 37 lines of Python from the credited upstream file 3317.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind the Number of Possible Ways for an Event · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def numberOfWays(self, n: int, x: int, y: int) -> int:    MOD = 1_000_000_007     @functools.lru_cache(None)    def fact(i: int) -> int:      return 1 if i <= 1 else i * fact(i - 1) % MOD     @functools.lru_cache(None)    def inv(i: int) -> int:      return pow(i, MOD - 2, MOD)     @functools.lru_cache(None)    def nCk(n: int, k: int) -> int:      return fact(n) * inv(fact(k)) * inv(fact(n - k)) % MOD     @functools.lru_cache(None)    def stirling(n: int, k: int) -> int:      """      Returns the number of ways to partition a set of n objects into k      non-empty subsets.       https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind      """      if k == 0 or n < k:        return 0      if k == 1 or n == k:        return 1      return (k * stirling(n - 1, k) + stirling(n - 1, k - 1)) % MOD     # 1. Choose `k` stages from `x` stages.    # 2. Partition `n` performers into `k` stages.    # 3. Permute `k` stages.    # 4. Score `k` stages with score in the range [1, y], so y^k ways.    return sum(nCk(x, k) * stirling(n, k) * fact(k) * pow(y, k, MOD) % MOD               for k in range(1, min(n, x) + 1)) % MOD

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