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Problem solution · Python

Maximum Total Beauty of the Gardens

Maximum Total Beauty of the Gardens: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
49 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

Sorting and greedy selection

For Maximum Total Beauty of the Gardens, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 49 lines of Python from the credited upstream file 2234.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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 Total Beauty of the Gardens · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def maximumBeauty(      self,      flowers: list[int],      newFlowers: int,      target: int,      full: int,      partial: int,  ) -> int:    n = len(flowers)     # If a garden is already complete, clamp it to the target.    flowers = [min(flower, target) for flower in flowers]    flowers.sort()     # All gardens are complete, so nothing we can do.    if flowers[0] == target:      return n * full     # Having many new flowers maximizes the beauty value.    if newFlowers >= n * target - sum(flowers):      return max(n * full, (n - 1) * full + (target - 1) * partial)     ans = 0    leftFlowers = newFlowers    # cost[i] := the cost to make flowers[0..i] the same    cost = [0] * n     for i in range(1, n):      # Plant (flowers[i] - flowers[i - 1]) flowers for flowers[0..i - 1].      cost[i] = cost[i - 1] + i * (flowers[i] - flowers[i - 1])     i = n - 1  # flowers' index (flowers[i + 1..n) are complete)    while flowers[i] == target:      i -= 1     while leftFlowers >= 0:      # To maximize the minimum number of incomplete flowers, we find the first      # index j that we can't make flowers[0..j] equal to flowers[j], then we      # know we can make flowers[0..j - 1] equal to flowers[j - 1]. In the      # meantime, evenly increase each of them to seek a bigger minimum value.      j = min(i + 1, bisect_right(cost, leftFlowers))      minIncomplete = flowers[j - 1] + (leftFlowers - cost[j - 1]) // j      ans = max(ans, (n - 1 - i) * full + minIncomplete * partial)      leftFlowers -= max(0, target - flowers[i])      i -= 1     return ans

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