Team A Solution

.github/pull_request_template.md

@@ -3,14 +3,18 @@
 PLEASE FILL IN THE FOLLOWING!
 
 ## Full Name
-John Johnson
+Mathew Pellarin
 
 ## UWindsor Email
-john.johnson@uwindsor.ca
+pellarim@uwindsor.ca
 
 ## Application Form
-- [ ] I certify I have submitted the [application form](https://forms.office.com/r/R4A1JyB3Xf)
+- [x] I certify I have submitted the [application form](https://forms.office.com/r/R4A1JyB3Xf)
 
 ## Briefly explain how your solution works and how to run it 
 
-My solution...
+My solution first checks for any cases that might cause the program to timeout due to run time length. It first checks if there is enough occurences of the letters within the board for the given the string and if not fails it. It then check if the last character appears more times then the first and reverses the string if so. 
+
+A recursive function is called and this function first goes through prechecks to check if a letter is matched, the rows and coloumns are in range, if the letter was visted before or if the letter does not match the needed letter. If the letter is correct, add it to a correct set and check all adjecent letters for the next character calling itself recursivly. Once all letters are found return the restult
+
+The variables are hard coded for testing. To test other inputs, chnaging the board and word variable will allow for other testing results. There is also a folder with a script that can be run on codeleet

solution.py

@@ -0,0 +1,65 @@
+from typing import List
+
+def exist(board: List[List[str]], word: str) -> bool:
+    # get all lengths of input
+    rows, cols, word_len = len(board), len(board[0]), len(word)
+
+    # convert board to a string
+    board_str = ''.join([''.join(row) for row in board])
+
+    # check if there are enough occurences of each character in the word in the board
+    for char in word:
+        if word.count(char) > board_str.count(char):
+            return False
+
+    # check if the last char appears less times then the first char, reverse the string if so
+    if board_str.count(word[0]) < board_str.count(word[-1]):
+        word = word[::-1]
+
+    # create a visited set
+    seen = set()
+
+    # row and col are the current position of the board, n is the current position of the word
+    def check(row, col, n=0):
+        # check if the word exists
+        if n == word_len:
+            return True
+
+        # check if the row and col are in range
+        if row < 0 or row >= rows or col < 0 or col >= cols:
+            return False
+
+        # check if the current position is visited
+        if (row, col) in seen:
+            return False
+
+        # backtrack if the current position of the board is not equal to the current position of the word
+        if board[row][col] != word[n]:
+            return False
+
+        # mark the letter as visited
+        seen.add((row, col))
+
+        # check if the word is found in any direction            
+        found = check(row + 1, col, n + 1) or \
+            check(row - 1, col, n + 1) or \
+            check(row, col + 1, n + 1) or \
+            check(row, col - 1, n + 1)
+
+        # mark the letter as unvisited
+        seen.remove((row, col))
+
+        # return the result
+        return found
+
+    # check if the word exists in the board
+    return any(check(row, col) for row in range(rows) for col in range(cols))
+
+if __name__ == '__main__':
+    board = [
+    ['A', 'B', 'C', 'E'],
+    ['S', 'F', 'C', 'S'],
+    ['A', 'D', 'E', 'E']]
+    word = "ABCCED"
+
+    print(exist(board, word))