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Problem solution · C++

Length of the Longest Increasing Path

Length of the Longest Increasing Path: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Binary search
Source
walkccc LeetCode Solutions
Length
44 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

Binary search

For Length of the Longest Increasing Path, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 44 lines of C++ from the credited upstream file 3288.cpp.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

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 codeLength of the Longest Increasing Path · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int maxPathLength(vector<vector<int>>& coordinates, int k) {    const int xk = coordinates[k][0];    const int yk = coordinates[k][1];    vector<pair<int, int>> leftCoordinates;    vector<pair<int, int>> rightCoordinates;     for (const vector<int>& coordinate : coordinates) {      const int x = coordinate[0];      const int y = coordinate[1];      if (x < xk && y < yk)        leftCoordinates.emplace_back(x, y);      else if (x > xk && y > yk)        rightCoordinates.emplace_back(x, y);    }     return 1 + lengthOfLIS(leftCoordinates) + lengthOfLIS(rightCoordinates);  }  private:  // Similar to 300. Longest Increasing Subsequence  int lengthOfLIS(vector<pair<int, int>>& coordinates) {    ranges::sort(coordinates, ranges::less{},                 [](const pair<int, int>& coordinate) {      const auto& [x, y] = coordinate;      return pair<int, int>{x, -y};    });    // tails[i] := the minimum tail of all the increasing subsequences having    // length i + 1    vector<int> tails;    for (const auto& [_, y] : coordinates)      if (tails.empty() || y > tails.back())        tails.push_back(y);      else        tails[firstGreaterEqual(tails, y)] = y;    return tails.size();  }   int firstGreaterEqual(const vector<int>& arr, int target) {    return ranges::lower_bound(arr, target) - arr.begin();  }};

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