Check if a word exists in a grid or not
Given a 2D grid of characters and a word, the task is to check if that word exists in the grid or not. A word can be matched in 4 directions at any point.
The 4 directions are, Horizontally Left and Right, Vertically Up and Down.
Examples:
Input: grid[][] = {"axmy",
"bgdf",
"xeet",
"raks"};
Output: Yes
a x m y
b g d f
x e e t
r a k s
Input: grid[][] = {"axmy",
"brdf",
"xeet",
"rass"};
Output : No
Source: Microsoft Interview
Approach: The idea used here is described in the steps below:
- Check every cell, if the cell has first character, then recur one by one and try all 4 directions from that cell for a match.
- Mark the position in the grid as visited and recur in the 4 possible directions.
- After recurring, again mark the position as unvisited.
- Once all the letters in the word is matched, return true.
Below is the implementation of the above approach:
C++
// C++ program to check if the word// exists in the grid or not#include <bits/stdc++.h>using namespace std;#define r 4#define c 5 // Function to check if a word exists in a grid// starting from the first match in the grid// level: index till which pattern is matched// x, y: current position in 2D arraybool findmatch(char mat[r], string pat, int x, int y, int nrow, int ncol, int level){ int l = pat.length(); // Pattern matched if (level == l) return true; // Out of Boundary if (x < 0 || y < 0 || x >= nrow || y >= ncol) return false; // If grid matches with a letter while // recursion if (mat[x][y] == pat[level]) { // Marking this cell as visited char temp = mat[x][y]; mat[x][y] = '#'; // finding subpattern in 4 directions bool res = findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) | findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) | findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) | findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1); // marking this cell // as unvisited again mat[x][y] = temp; return res; } else // Not matching then false return false;} // Function to check if the word exists in the grid or notbool checkMatch(char mat[r], string pat, int nrow, int ncol){ int l = pat.length(); // if total characters in matrix is // less then pattern lenghth if (l > nrow * ncol) return false; // Traverse in the grid for (int i = 0; i < nrow; i++) { for (int j = 0; j < ncol; j++) { // If first letter matches, then recur and check if (mat[i][j] == pat[0]) if (findmatch(mat, pat, i, j, nrow, ncol, 0)) return true; } } return false;} // Driver Codeint main(){ char grid[r] = { "axmy", "bgdf", "xeet", "raks" }; // Function to check if word exists or not if (checkMatch(grid, "geeks", r, c)) cout << "Yes"; else cout << "No"; return 0; } |
Java
// Java program to check if the word// exists in the grid or notclass GFG{ static final int r = 4;static final int c = 4; // Function to check if a word exists in a grid// starting from the first match in the grid// level: index till which pattern is matched// x, y: current position in 2D arraystatic boolean findmatch(char mat[][], String pat, int x, int y, int nrow, int ncol, int level){ int l = pat.length(); // Pattern matched if (level == l) return true; // Out of Boundary if (x < 0 || y < 0 || x >= nrow || y >= ncol) return false; // If grid matches with a letter while // recursion if (mat[x][y] == pat.charAt(level)) { // Marking this cell as visited char temp = mat[x][y]; mat[x][y] = '#'; // finding subpattern in 4 directions boolean res = findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) | findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) | findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) | findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1); // marking this cell // as unvisited again mat[x][y] = temp; return res; } else // Not matching then false return false;} // Function to check if the word exists in the grid or notstatic boolean checkMatch(char mat[][], String pat, int nrow, int ncol){ int l = pat.length(); // if total characters in matrix is // less then pattern lenghth if (l > nrow * ncol) return false; // Traverse in the grid for (int i = 0; i < nrow; i++) { for (int j = 0; j < ncol; j++) { // If first letter matches, then recur and check if (mat[i][j] == pat.charAt(0)) if (findmatch(mat, pat, i, j, nrow, ncol, 0)) return true; } } return false;} // Driver Codepublic static void main(String[] args){ char grid[][] = { "axmy".toCharArray(), "bgdf".toCharArray(), "xeet".toCharArray(), "raks".toCharArray() }; // Function to check if word exists or not if (checkMatch(grid, "geeks", r, c)) System.out.print("Yes"); else System.out.print("No");}} // This code is contributed by 29AjayKumar |
Python3
# Python3 program to check if the word # exists in the grid or not r = 4c = 4 # Function to check if a word exists # in a grid starting from the first # match in the grid level: index till # which pattern is matched x, y: current # position in 2D array def findmatch(mat, pat, x, y, nrow, ncol, level) : l = len(pat) # Pattern matched if (level == l) : return True # Out of Boundary if (x < 0 or y < 0 or x >= nrow or y >= ncol) : return False # If grid matches with a letter # while recursion if (mat[x][y] == pat[level]) : # Marking this cell as visited temp = mat[x][y] mat[x].replace(mat[x][y], "#") # finding subpattern in 4 directions res = (findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) | findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) | findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) | findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1)) # marking this cell as unvisited again mat[x].replace(mat[x][y], temp) return res else : # Not matching then false return False # Function to check if the word# exists in the grid or not def checkMatch(mat, pat, nrow, ncol) : l = len(pat) # if total characters in matrix is # less then pattern lenghth if (l > nrow * ncol) : return False # Traverse in the grid for i in range(nrow) : for j in range(ncol) : # If first letter matches, then # recur and check if (mat[i][j] == pat[0]) : if (findmatch(mat, pat, i, j, nrow, ncol, 0)) : return True return False # Driver Code if __name__ == "__main__" : grid = ["axmy", "bgdf", "xeet", "raks"] # Function to check if word # exists or not if (checkMatch(grid, "geeks", r, c)) : print("Yes") else : print("No") # This code is contributed by Ryuga |
C#
// C# program to check if the word// exists in the grid or notusing System; class GFG{ static readonly int r = 4;static readonly int c = 4; // Function to check if a word exists in a grid// starting from the first match in the grid// level: index till which pattern is matched// x, y: current position in 2D arraystatic bool findmatch(char [,]mat, String pat, int x, int y, int nrow, int ncol, int level){ int l = pat.Length; // Pattern matched if (level == l) return true; // Out of Boundary if (x < 0 || y < 0 || x >= nrow || y >= ncol) return false; // If grid matches with a letter while // recursion if (mat[x, y] == pat[level]) { // Marking this cell as visited char temp = mat[x, y]; mat[x, y] = '#'; // finding subpattern in 4 directions bool res = findmatch(mat, pat, x - 1, y, nrow, ncol, level + 1) | findmatch(mat, pat, x + 1, y, nrow, ncol, level + 1) | findmatch(mat, pat, x, y - 1, nrow, ncol, level + 1) | findmatch(mat, pat, x, y + 1, nrow, ncol, level + 1); // marking this cell // as unvisited again mat[x, y] = temp; return res; } else // Not matching then false return false;} // Function to check if the word exists in the grid or notstatic bool checkMatch(char [,]mat, String pat, int nrow, int ncol){ int l = pat.Length; // if total characters in matrix is // less then pattern lenghth if (l > nrow * ncol) return false; // Traverse in the grid for (int i = 0; i < nrow; i++) { for (int j = 0; j < ncol; j++) { // If first letter matches, then recur and check if (mat[i, j] == pat[0]) if (findmatch(mat, pat, i, j, nrow, ncol, 0)) return true; } } return false;} // Driver Codepublic static void Main(String[] args){ char [,]grid = { {'a','x','m','y'}, {'b','g','d','f'}, {'x','e','e','t'}, {'r','a','k','s'} }; // Function to check if word exists or not if (checkMatch(grid, "geeks", r, c)) Console.Write("Yes"); else Console.Write("No");}} // This code is contributed by 29AjayKumar |
Output:
Yes
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