Algorithm (Math)

1. If there are only two pigs and 15mins test pereiod, then we can form a 5x5 matrix and use one pig to test the row and another one to test the column.
2. With 3 pigs we instead of a 5x5 matrix will get a 5x5x5 cube.
3. So the problem becomes solvable by finding the smallest $n$-dimensional cube whose volume is larger than buckets. This could be done either using binary search or just linear scan. We settle for linear scan.

Code

template <class F>
struct recursive {
  F f;
  template <class... Ts>
  decltype(auto) operator()(Ts&&... ts)  const { return f(std::ref(*this), std::forward<Ts>(ts)...); }

  template <class... Ts>
  decltype(auto) operator()(Ts&&... ts)  { return f(std::ref(*this), std::forward<Ts>(ts)...); }
};

template <class F> recursive(F) -> recursive<F>;
auto const rec = [](auto f){ return recursive{std::move(f)}; };

class Solution {
 public:
  int poorPigs(int buckets, int minutesToDie, int minutesToTest) {
    const int n = minutesToTest / minutesToDie;

    const int solution = rec([&](auto && f, int acc, int count) -> int {
      if (acc < buckets)
        return f(acc * (n + 1), count + 1);
      else
        return count;
    })(1, 0);

    return solution;
};