Word Search in a 2D Board

Write a C program to determine if a given word exists in a 2D board of characters. The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same cell cannot be used more than once.

Example 1

  
Input:
    

      
board = [
        ['A','B','C','E'],
        ['S','F','C','S'],
        ['A','D','E','E']
      ]
      
word = "ABCCED"
    
  
  
Output: true
  
Explanation: 
    

      
Step 1: Start at board[0][0] = 'A', which matches the first character of "ABCCED".
      
Step 2: Move right to board[0][1] = 'B', which matches the second character.
      
Step 3: Move right to board[0][2] = 'C', which matches the third character.
      
Step 4: Move down to board[1][2] = 'C', which matches the fourth character.
      
Step 5: Move down to board[2][2] = 'E', which matches the fifth character.
      
Step 6: Move left to board[2][1] = 'D', which matches the last character.
      
Step 7: The word "ABCCED" is found in the board.
    
  

Example 2

  
Input:
    

      
board = [
        ['A','B','C','E'],
        ['S','F','C','S'],
        ['A','D','E','E']
      ]
      
word = "SEE"
    
  
  
Output: true
  
Explanation: 
    

      
Step 1: One possible path starts at board[1][0] = 'S'.
      
Step 2: Move right to board[1][3] = 'S' (after trying other paths that don't work).
      
Step 3: Move up to board[0][3] = 'E'.
      
Step 4: Move down to board[2][3] = 'E'.
      
Step 5: The word "SEE" is found in the board.
    
  

Constraints

  
m == board.length
  
n == board[i].length
  
1 <= m, n <= 6
  
1 <= word.length <= 15
  
board and word consist of only uppercase English letters
  
Time Complexity: O(m*n*4^L), where L is the length of the word
  
Space Complexity: O(m*n) for the visited array and recursion stack