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Problem solution · C++

Maximize the Distance Between Points on a Square

Maximize the Distance Between Points on a Square: 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
89 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 the Distance Between Points on a Square, 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

  • 89 lines of C++ from the credited upstream file 3464.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 4 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 the Distance Between Points on a Square · C++C++
Use this to learn the idea, then write your own version.
struct Sequence {  int startX;  int startY;  int endX;  int endY;  int length;}; class Solution { public:  int maxDistance(int side, vector<vector<int>>& points, int k) {    const vector<pair<int, int>> ordered = getOrderedPoints(side, points);    int l = 0;    int r = side;     while (l < r) {      const int m = (l + r + 1) / 2;      if (isValidDistance(ordered, k, m))        l = m;      else        r = m - 1;    }     return l;  }  private:  // Returns true if we can select `k` points such that the minimum Manhattan  // distance between any two consecutive chosen points is at least `m`.  bool isValidDistance(const vector<pair<int, int>>& ordered, int k, int d) {    deque<Sequence> dq{{ordered[0].first, ordered[0].second, ordered[0].first,                        ordered[0].second, 1}};    int maxLength = 1;     for (int i = 1; i < ordered.size(); ++i) {      const auto& [x, y] = ordered[i];      int startX = x;      int startY = y;      int length = 1;      while (!dq.empty() &&             (abs(x - dq.front().endX) + abs(y - dq.front().endY) >= d)) {        if (abs(x - dq.front().startX) + abs(y - dq.front().startY) >= d &&            dq.front().length + 1 >= length) {          startX = dq.front().startX;          startY = dq.front().startY;          length = dq.front().length + 1;          maxLength = max(maxLength, length);        }        dq.pop_front();      }      dq.emplace_back(startX, startY, x, y, length);    }     return maxLength >= k;  }   // Returns the ordered points on the perimeter of a square of side length  // `side`, starting from left, top, right, and bottom boundaries.  vector<pair<int, int>> getOrderedPoints(int side,                                          vector<vector<int>>& points) {    vector<pair<int, int>> left;    vector<pair<int, int>> top;    vector<pair<int, int>> right;    vector<pair<int, int>> bottom;     for (const vector<int>& point : points) {      const int x = point[0];      const int y = point[1];      if (x == 0 && y > 0)        left.emplace_back(x, y);      else if (x > 0 && y == side)        top.emplace_back(x, y);      else if (x == side && y < side)        right.emplace_back(x, y);      else        bottom.emplace_back(x, y);    }     ranges::sort(left);    ranges::sort(top);    ranges::sort(right, greater<>());    ranges::sort(bottom, greater<>());    left.insert(left.end(), top.begin(), top.end());    left.insert(left.end(), right.begin(), right.end());    left.insert(left.end(), bottom.begin(), bottom.end());    return left;  }};

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