CPMagical CP Library

Problem solution · Python

Find Sum of Array Product of Magical Sequences

Find Sum of Array Product of Magical Sequences: 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
27 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 Sum of Array Product of Magical Sequences, 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

  • 27 lines of Python from the credited upstream file 3539.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 codeFind Sum of Array Product of Magical Sequences · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def magicalSum(self, m: int, k: int, nums: list[int]) -> int:    MOD = 1_000_000_007     @functools.lru_cache(None)    def dp(m: int, k: int, i: int, carry: int) -> int:      """      Returns the number of magical sequences of length `k` that can be formed      from the first `i` numbers in `nums` with at most `m` elements.      """      if m < 0 or k < 0 or (m + carry.bit_count() < k):        return 0      if m == 0:        return int(k == carry.bit_count())      if i == len(nums):        return 0      res = 0      for count in range(m + 1):        contribution = math.comb(m, count) * pow(nums[i], count, MOD) % MOD        newCarry = carry + count        res += dp(m - count, k - (newCarry % 2),                  i + 1, newCarry // 2) * contribution        res %= MOD      return res     return dp(m, k, 0, 0)

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗