Search a Word in a 2D Grid of characters
Given a 2D grid of characters and a word, find all occurrences of the given word in the grid. A word can be matched in all 8 directions at any point. Word is said to be found in a direction if all characters match in this direction (not in zig-zag form).
The 8 directions are, Horizontally Left, Horizontally Right, Vertically Up and 4 Diagonal directions.
Example:
Input: grid[][] = {"GEEKSFORGEEKS",
"GEEKSQUIZGEEK",
"IDEQAPRACTICE"};
word = "GEEKS"
Output: pattern found at 0, 0
pattern found at 0, 8
pattern found at 1, 0
Explanation: 'GEEKS' can be found as prefix of
1st 2 rows and suffix of first row
Input: grid[][] = {"GEEKSFORGEEKS",
"GEEKSQUIZGEEK",
"IDEQAPRACTICE"};
word = "EEE"
Output: pattern found at 0, 2
pattern found at 0, 10
pattern found at 2, 2
pattern found at 2, 12
Explanation: EEE can be found in first row
twice at index 2 and index 10
and in second row at 2 and 12
Below diagram shows a bigger grid and presence of different words in it.
Source: Microsoft Interview Question.
Approach: The idea used here is simple, we check every cell. If cell has first character, then we one by one try all 8 directions from that cell for a match. Implementation is interesting though. We use two arrays x[] and y[] to find next move in all 8 directions.
Below are implementation of the same:
C++
// C++ programs to search a word in a 2D grid #include <bits/stdc++.h> using namespace std; // Rows and columns in given grid #define R 3 #define C 14 // For searching in all 8 direction int x[] = { -1, -1, -1, 0, 0, 1, 1, 1 }; int y[] = { -1, 0, 1, -1, 1, -1, 0, 1 }; // This function searches in // all 8-direction from point // (row, col) in grid[][] bool search2D(char grid[R][C], int row, int col, string word) { // If first character of word doesn't // match with given starting point in grid. if (grid[row][col] != word[0]) return false; int len = word.length(); // Search word in all 8 directions // starting from (row, col) for (int dir = 0; dir < 8; dir++) { // Initialize starting point // for current direction int k, rd = row + x[dir], cd = col + y[dir]; // First character is already checked, // match remaining characters for (k = 1; k < len; k++) { // If out of bound break if (rd >= R || rd < 0 || cd >= C || cd < 0) break; // If not matched, break if (grid[rd][cd] != word[k]) break; // Moving in particular direction rd += x[dir], cd += y[dir]; } // If all character matched, then value of must // be equal to length of word if (k == len) return true; } return false; } // Searches given word in a given // matrix in all 8 directions void patternSearch(char grid[R][C], string word) { // Consider every point as starting // point and search given word for (int row = 0; row < R; row++) for (int col = 0; col < C; col++) if (search2D(grid, row, col, word)) cout << "pattern found at " << row << ", " << col << endl; } // Driver program int main() { char grid[R][C] = { "GEEKSFORGEEKS", "GEEKSQUIZGEEK", "IDEQAPRACTICE" }; patternSearch(grid, "GEEKS"); cout << endl; patternSearch(grid, "EEE"); return 0; } |
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Java
// Java program to search // a word in a 2D grid import java.io.*; import java.util.*; class GFG { // Rows and columns in the given grid static int R, C; // For searching in all 8 direction static int[] x = { -1, -1, -1, 0, 0, 1, 1, 1 }; static int[] y = { -1, 0, 1, -1, 1, -1, 0, 1 }; // This function searches in all // 8-direction from point // (row, col) in grid[][] static boolean search2D(char[][] grid, int row, int col, String word) { // If first character of word // doesn't match with // given starting point in grid. if (grid[row][col] != word.charAt(0)) return false; int len = word.length(); // Search word in all 8 directions // starting from (row, col) for (int dir = 0; dir < 8; dir++) { // Initialize starting point // for current direction int k, rd = row + x[dir], cd = col + y[dir]; // First character is already checked, // match remaining characters for (k = 1; k < len; k++) { // If out of bound break if (rd >= R || rd < 0 || cd >= C || cd < 0) break; // If not matched, break if (grid[rd][cd] != word.charAt(k)) break; // Moving in particular direction rd += x[dir]; cd += y[dir]; } // If all character matched, // then value of must // be equal to length of word if (k == len) return true; } return false; } // Searches given word in a given // matrix in all 8 directions static void patternSearch( char[][] grid, String word) { // Consider every point as starting // point and search given word for (int row = 0; row < R; row++) { for (int col = 0; col < C; col++) { if (search2D(grid, row, col, word)) System.out.println( "pattern found at " + row + ", " + col); } } } // Driver code public static void main(String args[]) { R = 3; C = 13; char[][] grid = { { 'G', 'E', 'E', 'K', 'S', 'F', 'O', 'R', 'G', 'E', 'E', 'K', 'S' }, { 'G', 'E', 'E', 'K', 'S', 'Q', 'U', 'I', 'Z', 'G', 'E', 'E', 'K' }, { 'I', 'D', 'E', 'Q', 'A', 'P', 'R', 'A', 'C', 'T', 'I', 'C', 'E' } }; patternSearch(grid, "GEEKS"); System.out.println(); patternSearch(grid, "EEE"); } } // This code is contributed by rachana soma |
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Python3
# Python3 program to search a word in a 2D grid class GFG: def __init__(self): self.R = None self.C = None self.dir = [[-1, 0], [1, 0], [1, 1], [1, -1], [-1, -1], [-1, 1], [0, 1], [0, -1]] # This function searches in all 8-direction # from point(row, col) in grid[][] def search2D(self, grid, row, col, word): # If first character of word doesn't match # with the given starting point in grid. if grid[row][col] != word[0]: return False # Search word in all 8 directions # starting from (row, col) for x, y in self.dir: # Initialize starting point # for current direction rd, cd = row + x, col + y flag = True # First character is already checked, # match remaining characters for k in range(1, len(word)): # If out of bound or not matched, break if (0 <= rd <self.R and 0 <= cd < self.C and word[k] == grid[rd][cd]): # Moving in particular direction rd += x cd += y else: flag = False break # If all character matched, then # value of flag must be false if flag: return True return False # Searches given word in a given matrix # in all 8 directions def patternSearch(self, grid, word): # Rows and columns in given grid self.R = len(grid) self.C = len(grid[0]) # Consider every point as starting point # and search given word for row in range(self.R): for col in range(self.C): if self.search2D(grid, row, col, word): print("pattern found at " + str(row) + ', ' + str(col)) # Driver Code if __name__=='__main__': grid = ["GEEKSFORGEEKS", "GEEKSQUIZGEEK", "IDEQAPRACTICE"] gfg = GFG() gfg.patternSearch(grid, 'GEEKS') print('') gfg.patternSearch(grid, 'EEE') # This code is contributed by Yezheng Li |
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C#
// C# program to search a word in a 2D grid using System; class GFG { // Rows and columns in given grid static int R, C; // For searching in all 8 direction static int[] x = { -1, -1, -1, 0, 0, 1, 1, 1 }; static int[] y = { -1, 0, 1, -1, 1, -1, 0, 1 }; // This function searches in all 8-direction // from point (row, col) in grid[, ] static bool search2D(char[, ] grid, int row, int col, String word) { // If first character of word doesn't match // with given starting point in grid. if (grid[row, col] != word[0]) { return false; } int len = word.Length; // Search word in all 8 directions // starting from (row, col) for (int dir = 0; dir < 8; dir++) { // Initialize starting point // for current direction int k, rd = row + x[dir], cd = col + y[dir]; // First character is already checked, // match remaining characters for (k = 1; k < len; k++) { // If out of bound break if (rd >= R || rd < 0 || cd >= C || cd < 0) { break; } // If not matched, break if (grid[rd, cd] != word[k]) { break; } // Moving in particular direction rd += x[dir]; cd += y[dir]; } // If all character matched, then value of k // must be equal to length of word if (k == len) { return true; } } return false; } // Searches given word in a given // matrix in all 8 directions static void patternSearch(char[, ] grid, String word) { // Consider every point as starting // point and search given word for (int row = 0; row < R; row++) { for (int col = 0; col < C; col++) { if (search2D(grid, row, col, word)) { Console.WriteLine("pattern found at " + row + ", " + col); } } } } // Driver code public static void Main(String[] args) { R = 3; C = 13; char[, ] grid = { { 'G', 'E', 'E', 'K', 'S', 'F', 'O', 'R', 'G', 'E', 'E', 'K', 'S' }, { 'G', 'E', 'E', 'K', 'S', 'Q', 'U', 'I', 'Z', 'G', 'E', 'E', 'K' }, { 'I', 'D', 'E', 'Q', 'A', 'P', 'R', 'A', 'C', 'T', 'I', 'C', 'E' } }; patternSearch(grid, "GEEKS"); Console.WriteLine(); patternSearch(grid, "EEE"); } } #This code is contributed by Rajput - Ji |
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Output:
pattern found at 0, 0 pattern found at 0, 8 pattern found at 1, 0 pattern found at 0, 2 pattern found at 0, 10 pattern found at 2, 2 pattern found at 2, 12
Complexity Analysis:
- Time complexity: O(R*C).
All the cells will be visited and traversed in all 8 directions, where R and C is side of matrix so time complexity is O(R*C). - Auxiliary Space: O(1).
As no extra space is needed.
Exercise: The above solution only print locations of word. Extend it to print the direction where word is present.
See this for solution of exercise.
This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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