From 5156c82ab402c0452f3dd97e0738f0b3cb000c1b Mon Sep 17 00:00:00 2001
From: Sahithipsl470 <sahithipsl470@gmail.com>
Date: Tue, 10 Mar 2026 19:52:31 -0400
Subject: [PATCH] "Graph-1 done"

---
 Problem-1.py | 25 +++++++++++++++++++++++++
 Problem-2.py | 35 +++++++++++++++++++++++++++++++++++
 2 files changed, 60 insertions(+)
 create mode 100644 Problem-1.py
 create mode 100644 Problem-2.py

diff --git a/Problem-1.py b/Problem-1.py
new file mode 100644
index 0000000..bafa782
--- /dev/null
+++ b/Problem-1.py
@@ -0,0 +1,25 @@
+# Time Complexity : O(n + t), t = number of trust pairs
+# Space Complexity : O(n), to track trust counts
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Explanation:
+# 1. Track two counts for each person: number of people they trust and number of people who trust them.
+# 2. The town judge trusts nobody (trusts count = 0) and is trusted by n-1 people.
+# 3. Scan all people and return the one satisfying these conditions. Return -1 if none.
+
+from typing import List
+
+class Solution:
+    def findJudge(self, n: int, trust: List[List[int]]) -> int:
+        trusts = [0] * (n + 1)  # number of people each person trusts
+        trusted_by = [0] * (n + 1)  # number of people who trust each person
+        
+        for a, b in trust:
+            trusts[a] += 1
+            trusted_by[b] += 1
+        
+        for i in range(1, n + 1):
+            if trusts[i] == 0 and trusted_by[i] == n - 1:
+                return i
+        return -1
\ No newline at end of file
diff --git a/Problem-2.py b/Problem-2.py
new file mode 100644
index 0000000..e2532ad
--- /dev/null
+++ b/Problem-2.py
@@ -0,0 +1,35 @@
+# Time Complexity : O(m * n), worst case visiting all cells
+# Space Complexity : O(m * n), visited matrix
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+
+# Explanation:
+# 1. Use DFS to explore the maze. From a cell, roll in each direction until hitting a wall.
+# 2. Only mark the stopping cell as visited and continue DFS from there.
+# 3. If we reach the destination, return True. Otherwise, continue exploring other directions.
+# 4. Return False if all paths have been explored.
+
+from typing import List
+
+class Solution:
+    def hasPath(self, maze: List[List[int]], start: List[int], destination: List[int]) -> bool:
+        m, n = len(maze), len(maze[0])
+        visited = [[False] * n for _ in range(m)]
+        
+        def dfs(x, y):
+            if [x, y] == destination:
+                return True
+            if visited[x][y]:
+                return False
+            visited[x][y] = True
+            for dx, dy in [(0,1),(1,0),(0,-1),(-1,0)]:
+                nx, ny = x, y
+                # roll until hitting wall
+                while 0 <= nx + dx < m and 0 <= ny + dy < n and maze[nx + dx][ny + dy] == 0:
+                    nx += dx
+                    ny += dy
+                if dfs(nx, ny):
+                    return True
+            return False
+        
+        return dfs(start[0], start[1])
\ No newline at end of file