LeetCode #1923 — HARD

Longest Common Subpath

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

There is a country of n cities numbered from 0 to n - 1. In this country, there is a road connecting every pair of cities.

There are m friends numbered from 0 to m - 1 who are traveling through the country. Each one of them will take a path consisting of some cities. Each path is represented by an integer array that contains the visited cities in order. The path may contain a city more than once, but the same city will not be listed consecutively.

Given an integer n and a 2D integer array paths where paths[i] is an integer array representing the path of the ith friend, return the length of the longest common subpath that is shared by every friend's path, or 0 if there is no common subpath at all.

A subpath of a path is a contiguous sequence of cities within that path.

Example 1:

Input: n = 5, paths = [[0,1,2,3,4],
                       [2,3,4],
                       [4,0,1,2,3]]
Output: 2
Explanation: The longest common subpath is [2,3].

Example 2:

Input: n = 3, paths = [[0],[1],[2]]
Output: 0
Explanation: There is no common subpath shared by the three paths.

Example 3:

Input: n = 5, paths = [[0,1,2,3,4],
                       [4,3,2,1,0]]
Output: 1
Explanation: The possible longest common subpaths are [0], [1], [2], [3], and [4]. All have a length of 1.

Constraints:

  • 1 <= n <= 105
  • m == paths.length
  • 2 <= m <= 105
  • sum(paths[i].length) <= 105
  • 0 <= paths[i][j] < n
  • The same city is not listed multiple times consecutively in paths[i].
Patterns Used
🔍Modified Binary Search→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: There is a country of n cities numbered from 0 to n - 1. In this country, there is a road connecting every pair of cities. There are m friends numbered from 0 to m - 1 who are traveling through the country. Each one of them will take a path consisting of some cities. Each path is represented by an integer array that contains the visited cities in order. The path may contain a city more than once, but the same city will not be listed consecutively. Given an integer n and a 2D integer array paths where paths[i] is an integer array representing the path of the ith friend, return the length of the longest common subpath that is shared by every friend's path, or 0 if there is no common subpath at all. A subpath of a path is a contiguous sequence of cities within that path.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

5
[[0,1,2,3,4],[2,3,4],[4,0,1,2,3]]

Example 2

3
[[0],[1],[2]]

Example 3

5
[[0,1,2,3,4],[4,3,2,1,0]]

Related Problems

  • Reconstruct Itinerary (reconstruct-itinerary)
  • Maximum Length of Repeated Subarray (maximum-length-of-repeated-subarray)
Step 02

Core Insight

What unlocks the optimal approach

  • If there is a common path with length x, there is for sure a common path of length y where y < x.
  • We can use binary search over the answer with the range [0, min(path[i].length)].
  • Using binary search, we want to verify if we have a common path of length m. We can achieve this using hashing.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1923: Longest Common Subpath
class Solution {
    int N = 100010;
    long[] h = new long[N];
    long[] p = new long[N];
    private int[][] paths;
    Map<Long, Integer> cnt = new HashMap<>();
    Map<Long, Integer> inner = new HashMap<>();

    public int longestCommonSubpath(int n, int[][] paths) {
        int left = 0, right = N;
        for (int[] path : paths) {
            right = Math.min(right, path.length);
        }
        this.paths = paths;
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            if (check(mid)) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }

    private boolean check(int mid) {
        cnt.clear();
        inner.clear();
        p[0] = 1;
        for (int j = 0; j < paths.length; ++j) {
            int n = paths[j].length;
            for (int i = 1; i <= n; ++i) {
                p[i] = p[i - 1] * 133331;
                h[i] = h[i - 1] * 133331 + paths[j][i - 1];
            }
            for (int i = mid; i <= n; ++i) {
                long val = get(i - mid + 1, i);
                if (!inner.containsKey(val) || inner.get(val) != j) {
                    inner.put(val, j);
                    cnt.put(val, cnt.getOrDefault(val, 0) + 1);
                }
            }
        }
        int max = 0;
        for (int val : cnt.values()) {
            max = Math.max(max, val);
        }
        return max == paths.length;
    }

    private long get(int l, int r) {
        return h[r] - h[l - 1] * p[r - l + 1];
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Related Problems
#74Search a 2D Matrixmedium→#81Search in Rotated Sorted Array IImedium→#153Find Minimum in Rotated Sorted Arraymedium→#154Find Minimum in Rotated Sorted Array IIhard→#162Find Peak Elementmedium→