Merge pull request #289 from satyam878/satyam1

wordCount

Author: SarthakKeshari

LeetCode/wordCount.java

@@ -0,0 +1,45 @@
+/*Given an m x n grid of characters board and a string word, return true if word exists in the grid.
+The word can be constructed from letters of sequentially adjacent cells,
+ where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
+ */
+
+class Solution {
+    public boolean exist(char[][] board, String word) {
+        
+        int M=board.length, N=board[0].length; //Finding out no of rows and column in matrix
+        
+        for(int i=0;i<M;i++)
+        {
+            for(int j=0;j<N;j++)
+            {
+                if(board[i][j]==word.charAt(0)) //comparing each element of matrix with first letter of word
+                {
+                    if(dfs(board,i,j,word,0)) //applying dfs and checking if the word is present or not
+                        return true;
+                }   
+            }
+        }
+        return false;
+    }
+    
+    public boolean dfs(char[][] board,int i,int j,String word,int pos)
+    {
+        if(pos==word.length()) //if position of word is equal to length of word that means word is present in array
+            return true;
+        
+       int M=board.length, N=board[0].length; 
+            if(i < 0 || i >= M || j < 0 || j >= N || board[i][j]!=word.charAt(pos) || board[i][j]=='#') //checking the necessary condition to find out whether word exists or not
+            return false;
+        
+        char dup=board[i][j];
+        board[i][j]='#';
+        
+        boolean ans =  dfs(board,  i + 1, j, word, pos + 1) ||   //traversing one letter right
+                            dfs(board, i - 1, j, word, pos + 1) ||  ////traversing one letter left
+                                dfs(board, i, j + 1, word, pos + 1) ||  //traversing one letter up
+                                    dfs(board,  i, j - 1, word, pos + 1);  ////traversing one letter down
+        
+        board[i][j]=dup; //backtracking the string 
+            return ans;
+    }
+}