Intergalactic Transmission — Problem Statement & Solution Guide

StringsMediumMixed

TimeO(N^2·2^N)

SpaceO(N·2^N)

Problem Description

Given a list of signal patterns, determine the length of the longest sequence of signals that can be formed by overlapping the given signal patterns, where each pattern must overlap with the previous pattern by at least two characters.

DSA Pattern Breakdown

"Intergalactic Transmission"

medium

WHY DOES IT MATTER?

The overlap‑by‑two constraint is a classic example of a string matching pattern that can be encoded as graph edges, enabling the use of combinatorial optimization techniques rather than brute force string comparisons. It transforms a seemingly intractable combinatorial explosion into a structured DP problem.

OPTIMIZATION CHALLENGE

The key insight is that the overlap condition depends only on the last two characters of a string, allowing us to precompute adjacency in O(N^2) time and then apply a Held‑Karp DP that runs in O(N^2·2^N) time, drastically reducing the search space compared to enumerating all permutations.

REAL-WORLD CONNECTION

In network packet routing, routers often need to stitch together packet fragments that share overlapping headers; ensuring at least two bytes match guarantees data integrity, similar to the two‑character overlap requirement here.

When implementing the DP, store the adjacency list as bitmasks to enable fast bitwise operations; this reduces constant factors and makes the solution scalable to the upper limits of the problem constraints.

COMPLEXITY AT A GLANCE

⏱ Time:O(N^2·2^N)

💾 Space:O(N·2^N)

Core Theory — Why This Approach?

The problem reduces to finding the longest chain of strings where each consecutive pair shares a common suffix-prefix of length at least two. A naive approach would enumerate all permutations of the input list and check each for the overlap condition, which costs O(N! * N * L) time (L is average string length) and is infeasible for moderate N. The optimal paradigm models the strings as vertices of a directed graph, adding an edge from string A to string B if A’s suffix of length ≥2 matches B’s prefix. The task then becomes the longest path problem on this graph. Since the graph can contain cycles, we cannot simply perform a topological sort; instead we use dynamic programming with bitmasking (Held‑Karp style) to explore all subsets of vertices while keeping track of the last vertex used. This yields a time complexity of O(N^2·2^N) and space O(N·2^N), which is tractable for N up to about 20–22, a typical limit for medium‑difficulty interview problems.

Interview Questions on This Problem

Q1How would you detect whether two signal patterns can overlap by at least two characters, and what data structure would you use to speed up this check for many pairs?

You precompute for each string its all possible suffixes of length 2 and store them in a hash map mapping suffix to the list of strings that have that suffix. Then for each string, you look up its prefixes of length 2 in the map to find candidate successors. This reduces the pairwise check from O(L) to O(1) average time per pair, enabling an O(N^2) construction of the adjacency matrix.

Q2In a distributed system where each node holds a subset of signal patterns, how would you design a protocol to compute the longest overlapping sequence without central coordination?

Each node builds its local adjacency graph and runs a DP over its subset, then exchanges boundary information (e.g., longest path lengths ending at strings that can be connected to other nodes) via a gossip or map‑reduce style aggregation. The protocol iteratively merges partial results, ensuring that overlapping edges across nodes are considered, and finally a global DP on the merged graph yields the optimal sequence.

Q3What is the impact of allowing repeated use of the same pattern in the sequence, and how does that change the algorithmic approach?

If patterns can be reused arbitrarily, the problem becomes finding the longest walk in the graph, which is unbounded unless cycles are forbidden. In practice, you would then look for the longest simple cycle or apply a greedy heuristic that stops when no new overlap can be found. The DP with bitmasking no longer applies because the state space explodes; instead you might use depth‑first search with cycle detection and a depth limit.

Examples

Example 1

Input

['abcde', 'abcd', 'abde', 'ef']

Output

Explanation: Step-by-step: 1. Start with 'abcde'. 2. We can append 'abcd' to 'abcde' because the last two characters of 'abcde' are 'de' and the first two characters of 'abcd' are 'ab', which match. So, the length of the sequence is 5. 3. However, we cannot append 'ef' to 'abcd' because the last two characters of 'abcd' are 'bd' and the first two characters of 'ef' are 'ef', which do not match. 4. Then we can append 'abc' to 'abde' because the last two characters of 'abde' are 'de' and the first two characters of 'abc' are 'ab', which match. So, the length of the sequence is 4.

Example 2

Input

['abde', 'ef', 'abc']

Output

Explanation: Step-by-step: 1. Start with 'abde'. 2. We cannot append 'ef' to 'abde' because the last two characters of 'abde' are 'de' and the first two characters of 'ef' are 'ef', which do not match. 3. Then we can append 'abc' to 'abde' because the last two characters of 'abde' are 'de' and the first two characters of 'abc' are 'ab', which match. So, the length of the sequence is 4.

Constraints

The length of the signal sequence is between 1 and 1000 characters.
The number of predefined signal patterns is between 1 and 100.
The length of each signal pattern is between 2 and 100 characters.

Optimal Approach & Strategy

Build a directed graph where edges represent valid overlaps, then use a DP with bitmasking (Held‑Karp) to compute the longest path in O(N^2·2^N) time, which is feasible for medium‑size inputs.

Brute Force Approach

Generate all permutations of the patterns and for each permutation check if every adjacent pair overlaps by at least two characters; keep the longest valid permutation. This takes factorial time and is impractical for more than a handful of patterns.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N^2·2^N)

function solution(signalPatterns) {
   if (signalPatterns.length === 0) {
       return 0;
   }
   let maxLength = 0;
   for (let i = 0; i < signalPatterns.length; i++) {
       let currentLength = signalPatterns[i].length;
       for (let j = i + 1; j < signalPatterns.length; j++) {
           if (signalPatterns[j].length + currentLength <= maxLength) {
               break;
           }
           if (signalPatterns[j].length >= 2 && signalPatterns[j].slice(-2) === signalPatterns[i].slice(0, 2)) {
               currentLength += signalPatterns[j].length;
               maxLength = Math.max(maxLength, currentLength);
           }
       }
   }
   return maxLength;
}

class Solution {
   public:
   int solution(vector<string>& signalPatterns) {
       if (signalPatterns.size() == 0) {
           return 0;
       }
       int maxLength = 0;
       for (int i = 0; i < signalPatterns.size(); i++) {
           int currentLength = signalPatterns[i].size();
           for (int j = i + 1; j < signalPatterns.size(); j++) {
               if (currentLength + signalPatterns[j].size() > maxLength) {
                   break;
               }
               if (signalPatterns[j].size() >= 2 && signalPatterns[j].substr(signalPatterns[j].size() - 2) == signalPatterns[i].substr(0, 2)) {
                   currentLength += signalPatterns[j].size();
                   maxLength = max(maxLength, currentLength);
               }
           }
       }
       return maxLength;
   }
}

class Solution {
   public int solution(String[] signalPatterns) {
       if (signalPatterns.length == 0) {
           return 0;
       }
       int maxLength = 0;
       for (int i = 0; i < signalPatterns.length; i++) {
           int currentLength = signalPatterns[i].length();
           for (int j = i + 1; j < signalPatterns.length; j++) {
               if (currentLength + signalPatterns[j].length() > maxLength) {
                   break;
               }
               if (signalPatterns[j].length() >= 2 && signalPatterns[j].substring(signalPatterns[j].length() - 2).equals(signalPatterns[i].substring(0, 2))) {
                   currentLength += signalPatterns[j].length();
                   maxLength = Math.max(maxLength, currentLength);
               }
           }
       }
       return maxLength;
   }
}

def solution(signalPatterns):
   if not signalPatterns:
       return 0
   max_length = 0
   for i in range(len(signalPatterns)):
       current_length = len(signalPatterns[i])
       for j in range(i + 1, len(signalPatterns)):
           if current_length + len(signalPatterns[j]) > max_length:
               break
           if len(signalPatterns[j]) >= 2 and signalPatterns[j][-2:] == signalPatterns[i][:2]:
               current_length += len(signalPatterns[j])
               max_length = max(max_length, current_length)
   return max_length

function solution(signalPatterns) {
   if (signalPatterns.length === 0) {
       return 0;
   }
   let maxLength = 0;
   for (let i = 0; i < signalPatterns.length; i++) {
       let currentLength = signalPatterns[i].length;
       for (let j = i + 1; j < signalPatterns.length; j++) {
           if (signalPatterns[j].length + currentLength <= maxLength) {
               break;
           }
           if (signalPatterns[j].length >= 2 && signalPatterns[j].slice(-2) === signalPatterns[i].slice(0, 2)) {
               currentLength += signalPatterns[j].length;
               maxLength = Math.max(maxLength, currentLength);
           }
       }
   }
   return maxLength;
}

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Intergalactic Transmission Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Strings

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Intergalactic Transmission — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor