Problem solution · C++

Maximum Product of Subsequences With an Alternating Sum Equal to K

Maximum Product of Subsequences With an Alternating Sum Equal to K: a C++ solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
43 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

Hash-based lookup

For Maximum Product of Subsequences With an Alternating Sum Equal to K, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 43 lines of C++ from the credited upstream file 3509.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • No explicit loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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.

Full codeMaximum Product of Subsequences With an Alternating Sum Equal to K · C++C++
Use this to learn the idea, then write your own version.
enum class State {  kFirst,     // first element - add to sum and start product  kSubtract,  // second element - subtract from sum and multiply product  kAdd        // third element - add to sum and multiply product}; class Solution { public:  int maxProduct(vector<int>& nums, int k, int limit) {    if (abs(k) > accumulate(nums.begin(), nums.end(), 0))      return -1;    unordered_map<string, int> mem;    const int ans = maxProduct(nums, 0, 1, State::kFirst, k, limit, mem);    return ans == kMin ? -1 : ans;  }  private:  static constexpr int kMin = -5000;   int maxProduct(const vector<int>& nums, int i, int product, State state,                 int k, int limit, unordered_map<string, int>& mem) {    if (i == nums.size())      return k == 0 && state != State::kFirst && product <= limit ? product                                                                  : kMin;    const string key = to_string(i) + "," + to_string(k) + "," +                       to_string(product) + "," +                       to_string(static_cast<int>(state));    if (mem.contains(key))      return mem[key];    int res = maxProduct(nums, i + 1, product, state, k, limit, mem);    if (state == State::kFirst)      res = max(res, maxProduct(nums, i + 1, nums[i], State::kSubtract,                                k - nums[i], limit, mem));    if (state == State::kSubtract)      res = max(res, maxProduct(nums, i + 1, min(product * nums[i], limit + 1),                                State::kAdd, k + nums[i], limit, mem));    if (state == State::kAdd)      res = max(res, maxProduct(nums, i + 1, min(product * nums[i], limit + 1),                                State::kSubtract, k - nums[i], limit, mem));    return mem[key] = res;  }}; 

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