Neon Grid Infiltration — Problem Statement & Solution Guide

HashingMediumMixed

TimeO(4^{R*C}) in the worst case (exponential), typically much lower with pruning

SpaceO(R*C) for the recursion stack and visited matrix

Problem Description

In the depths of a cyberpunk hacker network, a rogue AI has created a neon grid of nodes, each containing a unique encryption key. To infiltrate the grid, a hacker must navigate through the nodes, collecting keys to unlock subsequent nodes. The hacker can move in any direction (up, down, left, right), but each node can only be visited once. The goal is to find the longest possible path through the grid.

DSA Pattern Breakdown

"Neon Grid Infiltration"

medium

WHY DOES IT MATTER?

Finding the longest simple path is a classic example of exhaustive search with pruning, a pattern that appears in puzzles, game AI, and routing problems where optimality matters but the search space is exponential.

OPTIMIZATION CHALLENGE

The key insight is to cut off branches early by comparing the maximum possible remaining steps (total cells - visited) with the best length found so far; if even a perfect continuation cannot improve the answer, the branch is discarded.

REAL-WORLD CONNECTION

In distributed systems, tracing the longest dependency chain of microservices (without cycles) mirrors this problem: each service call is a node, and you need the longest acyclic call sequence to identify latency bottlenecks.

During an interview, start with a clean recursive template, add the visited matrix, and only then sprinkle in pruning. A small, correct DFS skeleton demonstrates systematic thinking; the pruning shows you understand performance constraints.

COMPLEXITY AT A GLANCE

⏱ Time:O(4^{R*C}) in the worst case (exponential), typically much lower with pruning

💾 Space:O(R*C) for the recursion stack and visited matrix

Core Theory — Why This Approach?

The Neon Grid Infiltration problem maps to finding the longest simple path in an undirected grid graph where vertices are cells and edges connect orthogonal neighbours. A simple path cannot revisit a vertex, which makes the problem combinatorial: the number of possible walks grows exponentially with the number of cells. A naïve brute‑force that enumerates every permutation of cells quickly becomes infeasible even for modest grids (e.g., a 5×5 grid already has >10^13 possible walks). The optimal paradigm for medium‑sized inputs is a depth‑first search with backtracking, treating the grid as a state‑space tree. Each recursive call marks the current cell as visited, explores the four neighbours, and then unmarks the cell on return. Pruning techniques—such as abandoning a branch when the remaining unvisited cells cannot possibly beat the best length found so far—keep the search tractable. For very small grids (≤15 cells) a bitmask DP (state = visited mask + current position) can guarantee optimality in O(2^{N}·N) time, but the DFS/backtrack approach is simpler and works well for typical interview constraints.

Interview Questions on This Problem

Q1How would you modify the algorithm to also return the actual longest path, not just its length?

Maintain a list (or string) representing the current traversal order. When a new maximum length is discovered, copy the current list into a global answer variable. Because the recursion backtracks, the list always reflects the current path, so copying it at the right moment yields the required path.

Q2If the grid contains obstacles (blocked cells), how does that affect the search space and what extra checks are needed?

Obstacles are simply treated as non‑existent vertices: before recursing into a neighbour, verify that the cell is within bounds, not blocked, and not already visited. This reduces the branching factor, often dramatically shrinking the search space, but the worst‑case exponential nature remains.

Q3Explain how you could use memoization with a bitmask to solve the problem for grids up to 5×5, and why this approach is not scalable beyond that.

Encode the visited cells as a bitmask of size R·C and store dp[mask][pos] = longest path length achievable from position pos with visited set mask. The recurrence tries all four neighbours that are unvisited, updating the mask. The state space is O(2^{R·C}·R·C), which is feasible for ≤25 cells (≈33 million states) but explodes for larger grids, making the method impractical beyond small dimensions.

Examples

Example 1

Input

grid = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Output

Longest path: [1, 2, 3, 6, 9, 8, 7, 4, 5]

Explanation: Step-by-step: with input grid, we start at node 1, then move right to node 2, then right to node 3. From node 3, we move down to node 6, then down to node 9. From node 9, we move left to node 8, then left to node 7, then up to node 4, and finally right to node 5, resulting in the longest possible path.

Example 2

Input

grid = [[1, 2], [3, 4]]

Output

Longest path: [1, 2, 4, 3]

Explanation: Step-by-step: with input grid, we start at node 1, then move right to node 2, then down to node 4. From node 4, we move left to node 3, resulting in the longest possible path.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9

Optimal Approach & Strategy

Run a DFS with backtracking, marking cells as visited, exploring four directions, and pruning branches that cannot improve the current best length.

Brute Force Approach

Enumerate every permutation of cells and check if each consecutive pair are neighbours, keeping the longest valid sequence.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(4^{R*C}) in the worst case (exponential), typically much lower with pruning

function solution(grid) { 
       let rows = grid.length; 
       let cols = grid[0].length; 
       let directions = [[-1, 0], [1, 0], [0, -1], [0, 1]]; 
       let maxPath = []; 
       function dfs(row, col, path) { 
           if (path.length > maxPath.length) { 
               maxPath = path.slice(); 
           } 
           for (let i = 0; i < directions.length; i++) { 
               let newRow = row + directions[i][0]; 
               let newCol = col + directions[i][1]; 
               if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols && !path.includes(grid[newRow][newCol])) { 
                   dfs(newRow, newCol, path.concat([grid[newRow][newCol]])); 
               } 
           } 
       } 
       for (let i = 0; i < rows; i++) { 
           for (let j = 0; j < cols; j++) { 
               dfs(i, j, [grid[i][j]]); 
           } 
       } 
       return maxPath; 
   }

class Solution { 
       public: 
           vector<int> solution(vector<vector<int>>& grid) { 
               int rows = grid.size(); 
               int cols = grid[0].size(); 
               vector<vector<int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; 
               vector<int> maxPath; 
               function<void(int, int, vector<int>)> dfs = [&](int row, int col, vector<int> path) { 
                   if (path.size() > maxPath.size()) { 
                       maxPath = path; 
                   } 
                   for (auto& dir : directions) { 
                       int newRow = row + dir[0]; 
                       int newCol = col + dir[1]; 
                       if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols && find(path.begin(), path.end(), grid[newRow][newCol]) == path.end()) { 
                           path.push_back(grid[newRow][newCol]); 
                           dfs(newRow, newCol, path); 
                           path.pop_back(); 
                       } 
                   } 
               }; 
               for (int i = 0; i < rows; i++) { 
                   for (int j = 0; j < cols; j++) { 
                       dfs(i, j, {grid[i][j]}); 
                   } 
               } 
               return maxPath; 
           } 
   }

class Solution { 
       public int[][] solution(int[][] grid) { 
           int rows = grid.length; 
           int cols = grid[0].length; 
           int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; 
           int[][] maxPath = new int[rows * cols][rows * cols]; 
           int maxPathLen = 0; 
           boolean[][] visited = new boolean[rows][cols]; 
           for (int i = 0; i < rows; i++) { 
               for (int j = 0; j < cols; j++) { 
                   dfs(grid, i, j, directions, visited, new ArrayList<>()); 
               } 
           } 
           return maxPath; 
       } 
       public void dfs(int[][] grid, int row, int col, int[][] directions, boolean[][] visited, ArrayList<Integer> path) { 
           if (path.size() > maxPathLen) { 
               maxPathLen = path.size(); 
           } 
           for (int i = 0; i < directions.length; i++) { 
               int newRow = row + directions[i][0]; 
               int newCol = col + directions[i][1]; 
               if (newRow >= 0 && newRow < grid.length && newCol >= 0 && newCol < grid[0].length && !visited[newRow][newCol]) { 
                   visited[newRow][newCol] = true; 
                   path.add(grid[newRow][newCol]); 
                   dfs(grid, newRow, newCol, directions, visited, path); 
                   path.remove(path.size() - 1); 
                   visited[newRow][newCol] = false; 
               } 
           } 
       } 
   }

def solution(grid): 
       rows, cols = len(grid), len(grid[0]) 
       directions = [[-1, 0], [1, 0], [0, -1], [0, 1]] 
       max_path = [] 
       def dfs(row, col, path): 
           nonlocal max_path 
           if len(path) > len(max_path): 
               max_path = path[:] 
           for i in range(len(directions)): 
               new_row, new_col = row + directions[i][0], col + directions[i][1] 
               if 0 <= new_row < rows and 0 <= new_col < cols and grid[new_row][new_col] not in path: 
                   dfs(new_row, new_col, path + [grid[new_row][new_col]]) 
       for i in range(rows): 
           for j in range(cols): 
               dfs(i, j, [grid[i][j]]) 
       return max_path

function solution(grid) { 
       let rows = grid.length; 
       let cols = grid[0].length; 
       let directions = [[-1, 0], [1, 0], [0, -1], [0, 1]]; 
       let maxPath = []; 
       function dfs(row, col, path) { 
           if (path.length > maxPath.length) { 
               maxPath = path.slice(); 
           } 
           for (let i = 0; i < directions.length; i++) { 
               let newRow = row + directions[i][0]; 
               let newCol = col + directions[i][1]; 
               if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols && !path.includes(grid[newRow][newCol])) { 
                   dfs(newRow, newCol, path.concat([grid[newRow][newCol]])); 
               } 
           } 
       } 
       for (let i = 0; i < rows; i++) { 
           for (let j = 0; j < cols; j++) { 
               dfs(i, j, [grid[i][j]]); 
           } 
       } 
       return maxPath; 
   }

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Neon Grid Infiltration Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Hashing

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Neon Grid Infiltration — 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