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Problem solution · Python

Pizza With 3n Slices

Pizza With 3n Slices: 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
25 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 Pizza With 3n Slices, 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

  • 25 lines of Python from the credited upstream file 1388.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 codePizza With 3n Slices · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def maxSizeSlices(self, slices: list[int]) -> int:    @functools.lru_cache(None)    def dp(i: int, j: int, k: int) -> int:      """      Returns the maximum the sum of slices if you can pick k slices from      slices[i..j).      """      if k == 1:        return max(slices[i:j])      # Note that j - i is not the number of all the left slices. Since you      # Might have chosen not to take a slice in a previous step, there would be      # Leftovers outside [i:j]. If you take slices[i], one of the slices your      # Friends take will be outside of [i:j], so the length of [i:j] is reduced      # By 2 instead of 3. Therefore, the minimum # Is 2 * k - 1 (the last step only      # Requires one slice).      if j - i < 2 * k - 1:        return -math.inf      return max(slices[i] + dp(i + 2, j, k - 1),                 dp(i + 1, j, k))     k = len(slices) // 3    return max(dp(0, len(slices) - 1, k),               dp(1, len(slices), k))

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