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Problem solution · C++

Find the Number of Possible Ways for an Event

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

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
71 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

Direct simulation

For Find the Number of Possible Ways for an Event, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 71 lines of C++ from the credited upstream file 3317.cpp.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int numberOfWays(int n, int x, int y) {    const int maxStages = min(n, x);    const auto [fact, invFact] = getFactAndInvFact(max(n, x));    const vector<vector<int>> stirling = getStirling(n, maxStages);    int ans = 0;     for (int k = 1; k <= maxStages; ++k) {      // 1. Choose `k` stages from `x` stages.      long events = nCk(x, k, fact, invFact);      // 2. Partition `n` performers into `k` stages.      events = events * stirling[n][k] % kMod;      // 3. Permute `k` stages.      events = events * fact[k] % kMod;      // 4. Score `k` stages with score in the range [1, y], so y^k ways.      events = events * modPow(y, k) % kMod;      ans = (ans + events) % kMod;    }     return ans;  }  private:  static constexpr int kMod = 1'000'000'007;   pair<vector<long>, vector<long>> getFactAndInvFact(int n) {    vector<long> fact(n + 1);    vector<long> invFact(n + 1);    vector<long> inv(n + 1);    fact[0] = invFact[0] = 1;    inv[0] = inv[1] = 1;    for (int i = 1; i <= n; ++i) {      if (i >= 2)        inv[i] = kMod - kMod / i * inv[kMod % i] % kMod;      fact[i] = fact[i - 1] * i % kMod;      invFact[i] = invFact[i - 1] * inv[i] % kMod;    }    return {fact, invFact};  }   int nCk(int n, int k, const vector<long>& fact, const vector<long>& invFact) {    return fact[n] * invFact[k] % kMod * invFact[n - k] % kMod;  }   // Returns a 2D array stirling, where stirling[i][j] := the number of ways to  // partition a set of i objects into j non-empty subsets.  //  // https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind  vector<vector<int>> getStirling(int n, int k) {    vector<vector<int>> stirling(n + 1, vector<int>(k + 1));    stirling[0][0] = 1;    for (int i = 1; i <= n; ++i) {      stirling[i][1] = 1;      for (int j = 2; j <= min(i, k); ++j)        stirling[i][j] = (static_cast<long>(j) * stirling[i - 1][j] +                          stirling[i - 1][j - 1]) %                         kMod;    }    return stirling;  }   long modPow(long x, long n) {    if (n == 0)      return 1;    if (n % 2 == 1)      return x * modPow(x % kMod, (n - 1)) % kMod;    return modPow(x * x % kMod, (n / 2)) % kMod;  }};

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