Search a Word in a 2D Grid of characters
Given a 2D grid of characters and a single word/an array of words, find all occurrences of the given word/words 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, Vertically Down 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 when a single word is given: 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; // 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, int row, int col, string word, int R, int C) { // If first character of word doesn't // match with given starting point in grid. if (*(grid+row*C+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*C+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 void patternSearch( char *grid, string word, int R, int C) { // 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, R, C)) cout << "pattern found at " << row << ", " << col << endl; } // Driver program int main() { int R = 3, C = 13; char grid[R][C] = { "GEEKSFORGEEKS" , "GEEKSQUIZGEEK" , "IDEQAPRACTICE" }; patternSearch(( char *)grid, "GEEKS" , R, C); cout << endl; patternSearch(( char *)grid, "EEE" , R, C); return 0; } |
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 (grid[row][col]==word.charAt( 0 ) && 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 |
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 |
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 |
Javascript
<script> // JavaScript program to search // a word in a 2D grid // Rows and columns in the given grid let R, C; // For searching in all 8 direction let x=[-1, -1, -1, 0, 0, 1, 1, 1]; let y=[-1, 0, 1, -1, 1, -1, 0, 1]; // This function searches in all // 8-direction from point // (row, col) in grid[][] function search2D(grid,row,col,word) { // If first character of word // doesn't match with // given starting point in grid. if (grid[row][col] != word[0]) return false ; let len = word.length; // Search word in all 8 directions // starting from (row, col) for (let dir = 0; dir < 8; dir++) { // Initialize starting point // for current direction let 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 function patternSearch( grid,word) { // Consider every point as starting // point and search given word for (let row = 0; row < R; row++) { for (let col = 0; col < C; col++) { if (search2D(grid, row, col, word)) document.write( "pattern found at " + row + ", " + col+ "<br>" ); } } } // Driver code R = 3; C = 13; let 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" ); document.write( "<br>" ); patternSearch(grid, "EEE" ); // This code is contributed by avanitrachhadiya2155 </script> |
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*8*len(str)).
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.
Approach when array of words is given: 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, put this check in a loop. We use mover array to store the manner in which next moves are possible={left,right,up,down,…diagonals}.
Below are implementation of the same:
C++
#include <bits/stdc++.h> using namespace std; // making a solution class to solve the problem and to keep // the components and functions of solution together class Solution { public : // making the possible moves in movers array vector<vector< int > > mover = { { 1, 0 }, { 0, 1 }, { -1, 0 }, { 0, -1 }, { 1, 1 }, { -1, -1 }, { 1, -1 }, { -1, 1 } }; // making the board global variable vector<vector< char > > board; // depth first search for the string, with the // coordinates and a visited array to take care that we // do not overlap the places visited already bool dfs( int x, int y, string& s, vector<vector< bool > > vis) { // if string length becomes 0 means string is found if (s.length() == 0) return true ; vis[x][y] = true ; // making a solution boolean to see if we can // perform depth search to find answer bool sol = false ; // making possible moves for ( int i = 0; i < mover.size(); i++) { int curr_x = mover[i][0] + x; int curr_y = mover[i][1] + y; // checking for out of bound areas if (curr_x >= 0 && curr_x < board.size()) { if (curr_y >= 0 && curr_y < board[0].size()) { // checking for similarity in the first // letter and the visited array if (board[curr_x][curr_y] == s[0] && vis[curr_x][curr_y] == false ) { string k = s.substr( 1); // removing the first letter // from the string sol |= dfs(curr_x, curr_y, k, vis); } } } } return sol; } // making a function findwords to find words along with // their location which inputs the board and list of // words vector<string> findWords(vector<vector< char > >& board, vector<string>& words) { this ->board = board; // making board a global variable vector<string> ans; vector<vector< bool > > vis( board.size(), vector< bool >(board[0].size(), false )); // visited array for ( auto & word : words) { for ( int i = 0; i < board.size(); i++) { for ( int j = 0; j < board[i].size(); j++) { if (board[i][j] == word[0]) { // if first letter of(i,j)== // string's first letter then we can // perform dfs to check the // possiblity of string being present // from location (i,j) string s = word.substr(1); if (dfs(i, j, s, vis)) { ans.push_back( word + "->{" + to_string(i) + "," + to_string(j) + "}" ); } } } if (ans.size() && ans.back() == word) break ; } } return ans; } }; int main() { // making 1 instance of class solution as solver Solution solver; vector<vector< char > > board = { { 'o' , 'a' , 'a' , 'n' }, { 'e' , 't' , 'a' , 'e' }, { 'i' , 'h' , 'k' , 'r' }, { 'i' , 'f' , 'l' , 'v' } }; vector<string> words = { "oath" , "pea" , "eat" , "rain" }; // using the function findwords from our solution class // to find the answer vector<string> ans = solver.findWords(board, words); // printing the answer for ( auto & part : ans) cout << part << endl; return 0; } //contributed by NamanAnand |
oath->{0,0} eat->{1,0} eat->{1,3}
Complexity Analysis:
- Time complexity: O(R*C*len(str)*Number(str)*len(str)).
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) for each string. - Auxiliary Space: O(R*C*Numberof(str)*len(str)). (due to visited array)
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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