Is The Most Similar Path in a Graph easy or hard?

The Most Similar Path in a Graph is classified as Hard because it combines graph traversal with dynamic programming. The difficulty comes from recognizing the correct state representation and implementing path reconstruction while keeping the transitions efficient.

The Most Similar Path in a Graph Python/Java solution

Most implementations follow the same DP pattern across languages. Build the adjacency list, initialize dp[0][v] based on the first target name, and update dp[i][v] by iterating over neighbors. The final path is reconstructed using stored parent pointers. This approach works consistently in Python, Java, C++, Go, and TypeScript.

How to solve The Most Similar Path in a Graph in O(m * E)?

Model the problem as layered dynamic programming. For each step i and city v, compute dp[i][v] as the minimum cost of reaching v from any neighbor at step i-1 plus the mismatch penalty with targetPath[i]. Iterate this transition for all m steps and store parent pointers to reconstruct the final path.

What is the best approach for The Most Similar Path in a Graph?

Dynamic programming on graph states is the best approach. Define dp[i][v] as the minimum mismatch cost when the path position is i and the current city is v. Transition from all neighboring cities and add a penalty if the city name differs from targetPath[i]. This runs in O(m * E) time and guarantees the optimal path.

Is The Most Similar Path in a Graph asked at Google/Amazon/Meta?

Graph dynamic programming and shortest-path style problems like this appear frequently in interviews at companies such as Google, Amazon, and Meta. The key skill tested is modeling a sequence alignment problem on a graph using DP states and transitions.

What data structure is used in The Most Similar Path in a Graph?

The main structures are an adjacency list to represent the graph and a 2D DP table dp[m][n] to track the minimum mismatch cost. A separate parent array is used to reconstruct the optimal path after computing the DP transitions.

What is the time complexity of The Most Similar Path in a Graph?

The optimal dynamic programming solution runs in O(m * E) time, where m is the length of targetPath and E is the number of edges in the graph. Each DP layer processes transitions across all edges. Space complexity is O(m * n) for storing DP values and parent pointers for path reconstruction.

Description Hints 2 Solution Video 25:20

1548. The Most Similar Path in a Graph

Hard59.2% acceptance1 companyPremium elsewhere, free here

We have n cities and m bi-directional roads where roads[i] = [a_i, b_i] connects city a_i with city b_i. Each city has a name consisting of exactly three upper-case English letters given in the string array names. Starting at any city x, you can reach any city y where y != x (i.e., the cities and the roads are forming an undirected connected graph).

You will be given a string array targetPath. You should find a path in the graph of the same length and with the minimum edit distance to targetPath.

You need to return the order of the nodes in the path with the minimum edit distance. The path should be of the same length of targetPath and should be valid (i.e., there should be a direct road between ans[i] and ans[i + 1]). If there are multiple answers return any one of them.

The edit distance is defined as follows:

Example 1:

Input: n = 5, roads = [[0,2],[0,3],[1,2],[1,3],[1,4],[2,4]], names = ["ATL","PEK","LAX","DXB","HND"], targetPath = ["ATL","DXB","HND","LAX"]
Output: [0,2,4,2]
Explanation: [0,2,4,2], [0,3,0,2] and [0,3,1,2] are accepted answers.
[0,2,4,2] is equivalent to ["ATL","LAX","HND","LAX"] which has edit distance = 1 with targetPath.
[0,3,0,2] is equivalent to ["ATL","DXB","ATL","LAX"] which has edit distance = 1 with targetPath.
[0,3,1,2] is equivalent to ["ATL","DXB","PEK","LAX"] which has edit distance = 1 with targetPath.

Example 2:

Input: n = 4, roads = [[1,0],[2,0],[3,0],[2,1],[3,1],[3,2]], names = ["ATL","PEK","LAX","DXB"], targetPath = ["ABC","DEF","GHI","JKL","MNO","PQR","STU","VWX"]
Output: [0,1,0,1,0,1,0,1]
Explanation: Any path in this graph has edit distance = 8 with targetPath.

Example 3:

Input: n = 6, roads = [[0,1],[1,2],[2,3],[3,4],[4,5]], names = ["ATL","PEK","LAX","ATL","DXB","HND"], targetPath = ["ATL","DXB","HND","DXB","ATL","LAX","PEK"]
Output: [3,4,5,4,3,2,1]
Explanation: [3,4,5,4,3,2,1] is the only path with edit distance = 0 with targetPath.
It's equivalent to ["ATL","DXB","HND","DXB","ATL","LAX","PEK"]

Constraints:

2 <= n <= 100
m == roads.length
n - 1 <= m <= (n * (n - 1) / 2)
0 <= a_i, b_i <= n - 1
a_i != b_i
The graph is guaranteed to be connected and each pair of nodes may have at most one direct road.
names.length == n
names[i].length == 3
names[i] consists of upper-case English letters.
There can be two cities with the same name.
1 <= targetPath.length <= 100
targetPath[i].length == 3
targetPath[i] consists of upper-case English letters.

Follow up: If each node can be visited only once in the path, What should you change in your solution?

Asked at 1 company

Google

Topics

Dynamic Programming Graph

Hints open one at a time

Hint 1

Create an array dp where dp[i][j] is the min edit distance for the path starting at node i and compared to index j of the targetPath.

Hint 2

Traverse the dp array to obtain a valid answer.

25:20

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Examples

Case 1Case 2Case 3

Input

5
[[0,2],[0,3],[1,2],[1,3],[1,4],[2,4]]
["ATL","PEK","LAX","DXB","HND"]
["ATL","DXB","HND","LAX"]

Input

4
[[1,0],[2,0],[3,0],[2,1],[3,1],[3,2]]
["ATL","PEK","LAX","DXB"]
["ABC","DEF","GHI","JKL","MNO","PQR","STU","VWX"]

Input

6
[[0,1],[1,2],[2,3],[3,4],[4,5]]
["ATL","PEK","LAX","ATL","DXB","HND"]
["ATL","DXB","HND","DXB","ATL","LAX","PEK"]

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