My solution

.github/pull_request_template.md

@@ -3,14 +3,18 @@
 PLEASE FILL IN THE FOLLOWING!
 
 ## Full Name
-John Johnson
+Derek Thibert
 
 ## UWindsor Email
-john.johnson@uwindsor.ca
+thibertd@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 uses a backtracking solution which is pretty much brute force which goes through every cell to find a letter that matches with the word. The solution starts with the first letter of the word (if found) and then scopes the neighbours for a match without using the same letter cell.
+
+To run it: make sure Python is installed and extensions are installed in Visual Studio Code. Click on the triangle in the upper right hand corner in Visual Studio Code and click on "Run Python File".
+
+You can also type "py solution.py" to run it.

solution.py

@@ -0,0 +1,35 @@
+from typing import List
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        ROWS, COLS = len(board), len(board[0])
+        path = set()
+
+        def dfs(r, c, i):
+            if i == len(word):
+                return True
+            if (r < 0 or c < 0 or
+                r >= ROWS or c >= COLS or
+                word[i] != board[r][c] or
+                (r, c) in path):
+                return False
+
+            path.add((r, c))
+            res = (dfs(r + 1, c, i + 1) or
+                   dfs(r - 1, c, i + 1) or
+                   dfs(r, c + 1, i + 1) or
+                   dfs(r, c - 1, i + 1))
+            path.remove((r, c))
+            return res
+
+        for r in range(ROWS):
+            for c in range(COLS):
+                if dfs(r, c, 0): return True
+        return False
+
+ob1 = Solution()
+print(ob1.exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]],"SEE"))
+ob2 = Solution()
+print(ob2.exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]],"ABCCED"))
+ob3 = Solution()
+print(ob3.exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]],"ABCB"))