Lexical Grid Exploration — Problem Statement & Solution Guide

BacktrackingMediumWord Search

TimeO(m·n·L·8)

SpaceO(m·n) + O(L)

Problem Description

Given a 2D grid of letters and a target string, find all occurrences of the target string in the grid. The target string can be found horizontally, vertically, or diagonally in the grid, in a contiguous path.

DSA Pattern Breakdown

"Lexical Grid Exploration"

medium

WHY DOES IT MATTER?

Backtracking is essential because it systematically explores all feasible paths while discarding invalid ones early, preventing combinatorial explosion. Without pruning, the algorithm would revisit the same states repeatedly, leading to exponential time.

OPTIMIZATION CHALLENGE

The pivotal insight is to stop exploring a path as soon as a character mismatch occurs, which reduces the branching factor from 8^L to a much smaller number in practice. Additionally, using a visited matrix ensures each cell is used at most once per path, preventing cycles.

REAL-WORLD CONNECTION

Consider a robot navigating a maze where each move corresponds to a direction in the grid. The robot must find a specific sequence of moves (the target string) without revisiting the same spot in a single attempt, mirroring the backtracking strategy used in the algorithm.

When explaining this to an interviewer, emphasize the importance of early pruning and the use of direction vectors. Show how the algorithm’s complexity scales with grid size and target length, and be ready to discuss edge cases like overlapping matches or single‑character targets.

COMPLEXITY AT A GLANCE

⏱ Time:O(m·n·L·8)

💾 Space:O(m·n) + O(L)

Core Theory — Why This Approach?

The problem of finding a target string in a 2D grid is a classic example of backtracking combined with depth‑first search (DFS). Each cell in the grid can serve as a potential starting point, and from there we recursively explore all eight possible directions (horizontal, vertical, and both diagonals) while matching the characters of the target string. The key to efficiency is pruning: as soon as a character mismatch occurs, the current path is abandoned, preventing exponential blow‑up that a naive exhaustive search would suffer from.

Naïve approaches typically iterate over every cell and then perform a brute‑force search that explores all possible paths of length equal to the target string, leading to a time complexity of O(m·n·8^L) where m and n are grid dimensions and L is the length of the target. This quickly becomes infeasible for moderate grid sizes or longer strings.

The optimal paradigm leverages DFS with backtracking and a visited matrix to avoid revisiting cells within the same path. By pruning mismatches early and limiting the search depth to the target length, the algorithm achieves a practical time complexity of O(m·n·L·8) and a space complexity of O(L) for the recursion stack plus O(m·n) for the visited matrix. This approach scales well and is the standard solution used in production systems that perform pattern matching on 2D data structures such as image processing or crossword puzzle solvers.

Interview Questions on This Problem

Q1How would you modify the algorithm if the target string could wrap around the edges of the grid?

To allow wrapping, treat the grid as a toroidal surface. When moving in any direction, compute the next cell using modulo arithmetic: nextRow = (currentRow + dRow + m) % m and nextCol = (currentCol + dCol + n) % n. The DFS logic remains the same, but the visited matrix must be reset for each new starting cell to avoid infinite loops.

Q2What changes would you make to support multiple target strings efficiently?

Build a Trie (prefix tree) containing all target strings. Perform a single DFS from each cell, traversing the Trie as you match characters. When a Trie node marks the end of a word, record a match. This reduces redundant work by sharing common prefixes and achieves a time complexity of O(m·n·L) where L is the maximum length among the targets.

Q3In a distributed system, how could you parallelize the search across multiple machines?

Partition the grid into overlapping subgrids that include a boundary of width equal to the target length minus one. Each worker processes its subgrid independently using the backtracking algorithm. After local results are collected, merge them while eliminating duplicates that arise from overlapping boundaries. This approach scales linearly with the number of workers while preserving correctness.

Examples

Example 1

Input

grid = [['A','B','C','E'],['S','F','C','S','E'],['A','D','E','E','S','O']]; target = 'ABCESEEESO';
return [[0,0],[0,1],[0,2],[1,0],[1,1],[1,2],[2,0],[2,1],[2,2]];

Output

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

Explanation: Step 1: Initialize the grid and target string. Step 2: Define a helper function to check if the target string can be found in the grid. Step 3: Iterate over each cell in the grid and call the helper function. Step 4: If the target string is found, add the coordinates to the result list. Step 5: Return the result list.

Example 2

Input

grid = [['A','B','C','D'],['S','F','C','S']]; target = 'ABCD';
return [[0,0],[0,1],[0,2],[1,0],[1,1]];

Output

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

Constraints

The grid will contain at least one row and one column.
The target string will be between 1 and 10 characters long.
All characters in the grid and target string will be uppercase letters.

Optimal Approach & Strategy

The optimal approach uses DFS with backtracking, starting from each cell and exploring only until a mismatch occurs, while marking visited cells to avoid cycles. This reduces the time complexity to O(m·n·L·8) and uses O(L) stack space plus O(m·n) for the visited matrix.

Brute Force Approach

A naive solution would iterate over every cell and then recursively explore all possible paths of length equal to the target string, checking each path for a match. This leads to an exponential time complexity of O(m·n·8^L).

Verified Code Solutions

Language:

JavaScript Solution

Time: O(m·n·L·8)

function lexicalGridExploration(grid, target) {
   const rows = grid.length;
   const cols = grid[0].length;
   const directions = [[0, 1], [1, 0], [0, -1], [-1, 0], [1, 1], [-1, -1], [1, -1], [-1, 1]];
   const result = [];

   function dfs(row, col, index) {
      if (index === target.length) {
         result.push([row, col]);
         return;
      }
      if (row < 0 || row >= rows || col < 0 || col >= cols || grid[row][col] !== target[index]) {
         return;
      }
      const temp = grid[row][col];
      grid[row][col] = '#';
      for (const [dr, dc] of directions) {
         dfs(row + dr, col + dc, index + 1);
      }
      grid[row][col] = temp;
   }

   for (let i = 0; i < rows; i++) {
      for (let j = 0; j < cols; j++) {
         dfs(i, j, 0);
      }
   }
   return result;
}

class Solution {
   public:
      vector<vector<int>> lexicalGridExploration(vector<vector<char>>& grid, string target) {
         int rows = grid.size();
         int cols = grid[0].size();
         vector<int> directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {-1, -1}, {1, -1}, {-1, 1}};
         vector<vector<int>> result;

         void dfs(int row, int col, int index) {
            if (index == target.size()) {
               result.push_back({row, col});
               return;
            }
            if (row < 0 || row >= rows || col < 0 || col >= cols || grid[row][col] != target[index]) {
               return;
            }
            char temp = grid[row][col];
            grid[row][col] = '#';
            for (auto& direction : directions) {
               dfs(row + direction[0], col + direction[1], index + 1);
            }
            grid[row][col] = temp;
         }

         for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
               dfs(i, j, 0);
            }
         }
         return result;
      }
   };

class Solution {
   public List<List<Integer>> lexicalGridExploration(char[][] grid, String target) {
      int rows = grid.length;
      int cols = grid[0].length;
      int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {-1, -1}, {1, -1}, {-1, 1}};
      List<List<Integer>> result = new ArrayList<>();

      public void dfs(int row, int col, int index) {
         if (index == target.length()) {
            result.add(Arrays.asList(row, col));
            return;
         }
         if (row < 0 || row >= rows || col < 0 || col >= cols || grid[row][col] != target.charAt(index)) {
            return;
         }
         char temp = grid[row][col];
         grid[row][col] = '#';
         for (int[] direction : directions) {
            dfs(row + direction[0], col + direction[1], index + 1);
         }
         grid[row][col] = temp;
      }

      for (int i = 0; i < rows; i++) {
         for (int j = 0; j < cols; j++) {
            dfs(i, j, 0);
         }
      }
      return result;
   }
}

def lexicalGridExploration(grid, target):
   rows, cols = len(grid), len(grid[0])
   directions = [[0, 1], [1, 0], [0, -1], [-1, 0], [1, 1], [-1, -1], [1, -1], [-1, 1]]
   result = []

   def dfs(row, col, index):
      if index == len(target):
         result.append([row, col])
         return
      if row < 0 or row >= rows or col < 0 or col >= cols or grid[row][col] != target[index]:
         return
      temp = grid[row][col]
      grid[row][col] = '#'
      for dr, dc in directions:
         dfs(row + dr, col + dc, index + 1)
      grid[row][col] = temp

   for i in range(rows):
      for j in range(cols):
         dfs(i, j, 0)
   return result

function lexicalGridExploration(grid, target) {
   const rows = grid.length;
   const cols = grid[0].length;
   const directions = [[0, 1], [1, 0], [0, -1], [-1, 0], [1, 1], [-1, -1], [1, -1], [-1, 1]];
   const result = [];

   function dfs(row, col, index) {
      if (index === target.length) {
         result.push([row, col]);
         return;
      }
      if (row < 0 || row >= rows || col < 0 || col >= cols || grid[row][col] !== target[index]) {
         return;
      }
      const temp = grid[row][col];
      grid[row][col] = '#';
      for (const [dr, dc] of directions) {
         dfs(row + dr, col + dc, index + 1);
      }
      grid[row][col] = temp;
   }

   for (let i = 0; i < rows; i++) {
      for (let j = 0; j < cols; j++) {
         dfs(i, j, 0);
      }
   }
   return result;
}

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Lexical Grid Exploration Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Backtracking

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Lexical Grid Exploration — 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

Solve in Interative Editor