diff --git a/ballinmaze.py b/ballinmaze.py
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index 0000000..1aa6e69
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+++ b/ballinmaze.py
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+# Time Complexity : O(N*M) where N * M is the size of the matrix
+# Space Complexity : O(N*M) as we are using a queue to store the positions 
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Your code here along with comments explaining your approach:
+# I am using BFS to solve this problem.
+# I am starting from the start position and exploring all possible paths.
+# I am using a queue to store the positions to be explored.
+# I am marking the visited positions by changing their value to 2.
+# I am doing this to save space instead of using a separate visited array.
+# I am exploring all four directions from each position and moving until I hit a wall.
+# If the final position is the destination, then return True.
+# If no path is found, return False.
+
+from typing import List
+from collections import deque
+class Solution:
+    def hasPath(self, maze: List[List[int]], start: List[int], destination: List[int]) -> bool:
+        m = len(maze)
+        n = len(maze[0])
+        dirs = [(0,1),(1,0),(-1,0),(0,-1)]
+        q = deque([(start[0], start[1])]) 
+        maze[start[0]][start[1]] = 2
+        while q:
+            r,c = q.popleft()
+            for dr,dc in dirs:
+                nr = r + dr
+                nc = c + dc
+                while nr < m and nr >= 0 and nc < n and nc >= 0 and maze[nr][nc] != 1:
+                    nr += dr
+                    nc += dc
+                nr -= dr
+                nc -= dc
+                if nr == destination[0] and nc == destination[1]:
+                    return True
+                if maze[nr][nc] != 2:
+                    q.append((nr,nc))
+                    maze[nr][nc] = 2
+        return False
+
+                
+
+        
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diff --git a/townjudge.py b/townjudge.py
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+# Time Complexity : O(N + t) where N is the number of people and t is the number of trust relationships
+# Space Complexity : O(N) as we are using two arrays of size N+1
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Your code here along with comments explaining your approach:
+# I am using two arrays to keep track of the in-degrees and out-degrees of each person.
+# I am iterating through the trust list and for each trust relationship, I am incrementing
+# the in-degree of the person being trusted and the out-degree of the person who trusts.
+# Finally, I am checking for a person who has in-degree of n-1 and out-degree of 0.
+# If such a person exists, I return their label, otherwise I return -1.
+
+from typing import List
+class Solution:
+    def findJudge(self, n: int, trust: List[List[int]]) -> int:
+        in_degree = [0] * (n+1)
+        out_degree = [0] * (n+1)
+        for i,j in trust:
+            in_degree[j] += 1
+            out_degree[i] += 1
+        for i in range(1,n+1):
+            if in_degree[i] == n -1 and out_degree[i] == 0:
+                return i
+        return -1
+        
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