CPMagical CP Library

Problem solution · C++

The Most Similar Path in a Graph

The Most Similar Path in a Graph: a C++ solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
67 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

Depth-first search

For The Most Similar Path in a Graph, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 67 lines of C++ from the credited upstream file 1548.cpp.
  • The implementation visibly relies on sequence storage.
  • 4 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeThe Most Similar Path in a Graph · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  vector<int> mostSimilar(int n, vector<vector<int>>& roads,                          vector<string>& names, vector<string>& targetPath) {    this->names = names;    this->targetPath = targetPath;    // cost[i][j] := the minimum cost to start from names[i] in path[j]    this->cost.resize(names.size(), vector<int>(targetPath.size(), -1));    // next[i][j] := the best next of names[i] in path[j]    this->next.resize(names.size(), vector<int>(targetPath.size()));    this->graph.resize(n);     for (const vector<int>& road : roads) {      graph[road[0]].push_back(road[1]);      graph[road[1]].push_back(road[0]);    }     int minDist = INT_MAX;    int start = 0;     for (int i = 0; i < n; ++i) {      const int dist = dfs(i, 0);      if (dist < minDist) {        minDist = dist;        start = i;      }    }     vector<int> ans;     while (ans.size() < targetPath.size()) {      ans.push_back(start);      start = next[start][ans.size() - 1];    }     return ans;  }  private:  vector<string> names;  vector<string> targetPath;  vector<vector<int>> cost;  vector<vector<int>> next;  vector<vector<int>> graph;   int dfs(int nameIndex, int pathIndex) {    if (cost[nameIndex][pathIndex] != -1)      return cost[nameIndex][pathIndex];     const int editDist = names[nameIndex] != targetPath[pathIndex];    if (pathIndex == targetPath.size() - 1)      return editDist;     int minDist = INT_MAX;     for (const int v : graph[nameIndex]) {      const int dist = dfs(v, pathIndex + 1);      if (dist < minDist) {        minDist = dist;        next[nameIndex][pathIndex] = v;      }    }     return cost[nameIndex][pathIndex] = editDist + minDist;  }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗