Completed Graph-1

Problem1.py

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+# Problem1 Find Judge (https://leetcode.com/problems/find-the-town-judge/)
+# Time Complexity: O(n + t),We loop through the trust list once (t edges), then loop through all n people once.
+# Space Complexity: O(n),We store an indegrees array of size n+1.
+# Approach:
+# The town judge is trusted by everyone else but trusts nobody.
+# For each trust pair [a, b], a trusts b, so we decrease a's score (a trusts someone, so a cannot be the judge)
+# and increase b's score (b is trusted by someone).
+# The judge is the one person whose final score equals n-1, meaning everyone else trusts them and they trust no one.
+
+class Solution:
+    def findJudge(self, n: int, trust: List[List[int]]) -> int:
+        indegrees = [0] * (n + 1)  # score for each person from 1 to n, index 0 is unused
+
+        for i in trust:  # go through each trust pair
+            indegrees[i[0]] -= 1  # person i[0] trusts someone, so they lose a point (cannot be judge if they trust anyone)
+            indegrees[i[1]] += 1  # person i[1]] is trusted by someone, so they gain a point
+
+        for i in range(1, n + 1):  # check every person from 1 to n
+            if indegrees[i] == n - 1:  # trusted by everyone else and trusts nobody
+                return i  # found the judge
+
+        return -1  # no one matches, no judge exists

Problem2.py

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+# Problem2 The Maze (https://leetcode.com/problems/the-maze/)
+# Time Complexity: O(m * n * max(m, n)),Each cell can be visited once as a "stopping point" -> O(m*n) DFS calls.From each stopping point, we roll in 4 directions, each roll can take up to O(max(m,n)) steps
+# Space Complexity: O(m * n),Recursion stack in the worst case can hold O(m*n) calls(the maze itself is reused as the visited array, so no extra visited grid is needed)
+# Approach:
+# The ball doesn't stop cell by cell, it rolls until it hits a wall or the border.
+# So from each position, roll in all 4 directions until blocked, and only stop there.
+# Treat each "stopping point" as a node, and DFS between stopping points.
+# Mark visited stopping points as -1 in the maze to avoid revisiting the same paths.
+
+class Solution:
+    def hasPath(self, maze, start, destination):
+        self.dirs = [(1,0),(0,1),(0,-1),(-1,0)]   # down, right, left, up
+        self.m = len(maze)                         # number of rows
+        self.n = len(maze[0])                       # number of columns
+
+        return self.dfs(maze, start[0], start[1], destination)
+
+    def dfs(self, maze, i, j, destination):
+        # base case 1: reached the destination
+        if destination[0] == i and destination[1] == j:
+            return True
+
+        # base case 2: already visited this stopping point, avoid infinite loop
+        if maze[i][j] == -1:
+            return False
+
+        maze[i][j] = -1   # mark current stopping point as visited
+
+        for dir in self.dirs:
+            r = dir[0] + i   # take one step in this direction
+            c = dir[1] + j
+
+            # keep rolling while still inside the maze and not hitting a wall
+            while r >= 0 and c >= 0 and r < self.m and c < self.n and maze[r][c] != 1:
+                r += dir[0]
+                c += dir[1]
+
+            # the loop above overshoots by one step (lands on wall/border), so step back
+            r -= dir[0]
+            c -= dir[1]
+
+            # recursively check if the destination is reachable from this new stopping point
+            if self.dfs(maze, r, c, destination):
+                return True
+
+        return False   # no direction led to the destination