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Problem solution · Python

Maximum Product of Subsequences With an Alternating Sum Equal to K

Maximum Product of Subsequences With an Alternating Sum Equal to K: 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
34 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 Maximum Product of Subsequences With an Alternating Sum Equal to K, 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

  • 34 lines of Python from the credited upstream file 3509.py.
  • The implementation visibly relies on sequence storage, cached states.
  • 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 codeMaximum Product of Subsequences With an Alternating Sum Equal to K · PythonPython
Use this to learn the idea, then write your own version.
from enum import Enum  class State(Enum):  FIRST = 0  # first element - add to sum and start product  SUBTRACT = 1  # second element - subtract from sum and multiply product  ADD = 2  # third element - add to sum and multiply product  class Solution:  def maxProduct(self, nums: list[int], k: int, limit: int) -> int:    MIN = -5000    if abs(k) > sum(nums):      return -1     @functools.lru_cache(None)    def dp(i: int, product: int, state: State, k: int) -> int:      if i == len(nums):        return (product if k == 0 and state != State.FIRST and product <= limit                else MIN)      res = dp(i + 1, product, state, k)      if state == State.FIRST:        res = max(res, dp(i + 1, nums[i], State.SUBTRACT, k - nums[i]))      if state == State.SUBTRACT:        res = max(res, dp(i + 1, min(product * nums[i], limit + 1),                          State.ADD, k + nums[i]))      if state == State.ADD:        res = max(res, dp(i + 1, min(product * nums[i], limit + 1),                          State.SUBTRACT, k - nums[i]))      return res     ans = dp(0, 1, State.FIRST, k)    return -1 if ans == MIN else ans

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