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Problem solution · C++

Maximize Active Section with Trade II

Maximize Active Section with Trade II: a C++ 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
116 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

  • 116 lines of C++ from the credited upstream file 3501.cpp.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 5 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 · C++C++
Use this to learn the idea, then write your own version.
#include <ranges> struct Group {  int start;  int length;}; class SparseTable { public:  SparseTable(const vector<int>& nums)      : n(nums.size()), st(std::bit_width(n) + 1, vector<int>(n + 1)) {    copy(nums.begin(), nums.end(), st[0].begin());    for (int i = 1; i <= bit_width(n); ++i)      for (int j = 0; j + (1 << i) <= n; ++j)        st[i][j] = max(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);  }   // Returns max(nums[l..r]).  int query(unsigned l, unsigned r) const {    const int i = bit_width(r - l + 1) - 1;    return max(st[i][l], st[i][r - (1 << i) + 1]);  }  private:  const unsigned n;  vector<vector<int>> st;  // st[i][j] := max(nums[j..j + 2^i - 1])}; class Solution { public:  vector<int> maxActiveSectionsAfterTrade(string s,                                          vector<vector<int>>& queries) {    const int n = s.length();    const int ones = ranges::count(s, '1');    const auto [zeroGroups, zeroGroupIndex] = getZeroGroups(s);    if (zeroGroups.empty())      return vector<int>(queries.size(), ones);     const SparseTable st(getZeroMergeLengths(zeroGroups));    vector<int> ans;     for (const vector<int>& query : queries) {      const int l = query[0];      const int r = query[1];      const int left = zeroGroupIndex[l] == -1                           ? -1                           : (zeroGroups[zeroGroupIndex[l]].length -                              (l - zeroGroups[zeroGroupIndex[l]].start));      const int right = zeroGroupIndex[r] == -1                            ? -1                            : (r - zeroGroups[zeroGroupIndex[r]].start + 1);      const auto [startAdjacentGroupIndex, endAdjacentGroupIndex] =          mapToAdjacentGroupIndices(              zeroGroupIndex[l] + 1,              s[r] == '1' ? zeroGroupIndex[r] : zeroGroupIndex[r] - 1);      int activeSections = ones;      if (s[l] == '0' && s[r] == '0' &&          zeroGroupIndex[l] + 1 == zeroGroupIndex[r])        activeSections = max(activeSections, ones + left + right);      else if (startAdjacentGroupIndex <= endAdjacentGroupIndex)        activeSections = max(            activeSections,            ones + st.query(startAdjacentGroupIndex, endAdjacentGroupIndex));      if (s[l] == '0' &&          zeroGroupIndex[l] + 1 <=              (s[r] == '1' ? zeroGroupIndex[r] : zeroGroupIndex[r] - 1))        activeSections =            max(activeSections,                ones + left + zeroGroups[zeroGroupIndex[l] + 1].length);      if (s[r] == '0' && zeroGroupIndex[l] < zeroGroupIndex[r] - 1)        activeSections =            max(activeSections,                ones + right + zeroGroups[zeroGroupIndex[r] - 1].length);      ans.push_back(activeSections);    }     return ans;  }  private:  // Returns the zero groups and the index of the zero group that contains the  // i-th character.  pair<vector<Group>, vector<int>> getZeroGroups(const string& s) {    vector<Group> zeroGroups;    vector<int> zeroGroupIndex;    for (int i = 0; i < s.length(); i++) {      if (s[i] == '0') {        if (i > 0 && s[i - 1] == '0')          ++zeroGroups.back().length;        else          zeroGroups.push_back({i, 1});      }      zeroGroupIndex.push_back(zeroGroups.size() - 1);    }    return {zeroGroups, zeroGroupIndex};  }   // Returns the sums of the lengths of the adjacent groups.  vector<int> getZeroMergeLengths(const vector<Group>& zeroGroups) {    vector<int> zeroMergeLengths;    for (const auto& [a, b] : zeroGroups | views::pairwise)      zeroMergeLengths.push_back(a.length + b.length);    return zeroMergeLengths;  }   // 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)  pair<int, int> mapToAdjacentGroupIndices(int startGroupIndex,                                           int endGroupIndex) {    return {startGroupIndex, endGroupIndex - 1};  }};

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