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Problem solution · Python

Maximize Active Section with Trade II

Maximize Active Section with Trade II: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
102 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

Sliding window or two pointers

For Maximize Active Section with Trade II, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 102 lines of Python from the credited upstream file 3501.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 codeMaximize Active Section with Trade II · PythonPython
Use this to learn the idea, then write your own version.
from dataclasses import dataclass  @dataclassclass Group:  start: int  length: int  class SparseTable:  def __init__(self, nums: list[int]):    self.n = len(nums)    # st[i][j] := max(nums[j..j + 2^i - 1])    self.st = [[0] * (self.n + 1) for _ in range(self.n.bit_length() + 1)]    self.st[0] = nums.copy()    for i in range(1, self.n.bit_length() + 1):      for j in range(self.n - (1 << i) + 1):        self.st[i][j] = max(            self.st[i - 1][j],            self.st[i - 1][j + (1 << (i - 1))])   def query(self, l: int, r: int) -> int:    """Returns max(nums[l..r])."""    i = (r - l + 1).bit_length() - 1    return max(self.st[i][l], self.st[i][r - (1 << i) + 1])  class Solution:  def maxActiveSectionsAfterTrade(      self,      s: str,      queries: list[list[int]]  ) -> list[int]:    ones = s.count('1')    zeroGroups, zeroGroupIndex = self._getZeroGroups(s)    if not zeroGroups:      return [ones] * len(queries)     st = SparseTable(self._getZeroMergeLengths(zeroGroups))     def getMaxActiveSections(l: int, r: int) -> int:      left = (-1 if zeroGroupIndex[l] == -1              else (zeroGroups[zeroGroupIndex[l]].length -                    (l - zeroGroups[zeroGroupIndex[l]].start)))      right = (-1 if zeroGroupIndex[r] == -1               else (r - zeroGroups[zeroGroupIndex[r]].start + 1))      startAdjacentGroupIndex, endAdjacentGroupIndex = self._mapToAdjacentGroupIndices(          zeroGroupIndex[l] + 1, zeroGroupIndex[r] if s[r] == '1' else zeroGroupIndex[r] - 1)      activeSections = ones      if (s[l] == '0' and s[r] == '0' and              zeroGroupIndex[l] + 1 == zeroGroupIndex[r]):        activeSections = max(activeSections, ones + left + right)      elif startAdjacentGroupIndex <= endAdjacentGroupIndex:        activeSections = max(            activeSections,            ones + st.query(startAdjacentGroupIndex, endAdjacentGroupIndex))      if (s[l] == '0' and          zeroGroupIndex[l] + 1 <= (zeroGroupIndex[r]                                    if s[r] == '1' else zeroGroupIndex[r] - 1)):        activeSections = max(activeSections, ones + left +                             zeroGroups[zeroGroupIndex[l] + 1].length)      if (s[r] == '0' and zeroGroupIndex[l] < zeroGroupIndex[r] - 1):        activeSections = max(activeSections, ones + right +                             zeroGroups[zeroGroupIndex[r] - 1].length)      return activeSections     return [getMaxActiveSections(l, r) for l, r in queries]   def _getZeroGroups(self, s: str) -> tuple[list[Group], list[int]]:    """    Returns the zero groups and the index of the zero group that contains the    i-th character.    """    zeroGroups = []    zeroGroupIndex = []    for i in range(len(s)):      if s[i] == '0':        if i > 0 and s[i - 1] == '0':          zeroGroups[-1].length += 1        else:          zeroGroups.append(Group(i, 1))      zeroGroupIndex.append(len(zeroGroups) - 1)    return zeroGroups, zeroGroupIndex   def _getZeroMergeLengths(self, zeroGroups: list[Group]) -> list[int]:    """Returns the sums of the lengths of the adjacent groups."""    return [a.length + b.length for a, b in itertools.pairwise(zeroGroups)]   def _mapToAdjacentGroupIndices(      self,      startGroupIndex: int,      endGroupIndex: int  ) -> tuple[int, int]:    """    Returns the indices of the adjacent groups that contain l and r completely.     e.g.    groupIndices = [0, 1, 2, 3]    adjacentGroupIndices = [0 (0, 1), 1 (1, 2), 2 (2, 3)]    map(startGroupIndex = 1, endGroupIndex = 3) -> (1, 2)    """    return startGroupIndex, endGroupIndex - 1

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