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.
- Define the state carried into one recursive or stack frame.
- Mark or choose the current state before exploring children.
- Combine child results or undo the choice when the branch finishes.
Code notes
- 49 lines of Python from the credited upstream file 1548.py.
- The implementation visibly relies on sequence storage.
- No explicit 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.
Use this to learn the idea, then write your own version.
1class Solution:2 def mostSimilar(self, n: int, roads: list[list[int]], names: list[str],3 targetPath: list[str]) -> list[int]:4 5 cost = [[-1] * len(targetPath) for _ in range(len(names))]6 7 next = [[0] * len(targetPath) for _ in range(len(names))]8 graph = [[] for _ in range(n)]9 10 for u, v in roads:11 graph[u].append(v)12 graph[v].append(u)13 14 minDist = math.inf15 start = 016 17 def dfs(nameIndex: int, pathIndex: int) -> int:18 if cost[nameIndex][pathIndex] != -1:19 return cost[nameIndex][pathIndex]20 21 editDist = names[nameIndex] != targetPath[pathIndex]22 if pathIndex == len(targetPath) - 1:23 return editDist24 25 minDist = math.inf26 27 for v in graph[nameIndex]:28 dist = dfs(v, pathIndex + 1)29 if dist < minDist:30 minDist = dist31 next[nameIndex][pathIndex] = v32 33 cost[nameIndex][pathIndex] = editDist + minDist34 return editDist + minDist35 36 for i in range(n):37 dist = dfs(i, 0)38 if dist < minDist:39 minDist = dist40 start = i41 42 ans = []43 44 while len(ans) < len(targetPath):45 ans.append(start)46 start = next[start][len(ans) - 1]47 48 return ans49