Done Graph-1

ball_in_maze.java

@@ -0,0 +1,61 @@
+// Time Complexity : O(m * n) because each stopping point is visited at most once
+// Space Complexity : O(m * n) for recursion stack in the worst case
+// 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 in three sentences only
+// We use DFS to explore all reachable stopping points in the maze.
+// From each position, the ball rolls in a direction until it hits a wall, and only the stopping position is explored further.
+// We mark visited stopping points to avoid revisiting the same state and prevent infinite loops.
+
+class Solution {
+    int[][] dirs;
+    int m;
+    int n;
+
+    public boolean hasPath(int[][] maze, int[] start, int[] destination) {
+        this.dirs = new int[][]{{-1,0},{1,0},{0,1},{0,-1}};
+        this.m = maze.length;
+        this.n = maze[0].length;
+
+        return dfs(maze, start[0], start[1], destination);
+    }
+
+    private boolean dfs(int[][] maze, int i, int j, int[] destination) {
+
+        // destination reached
+        if(destination[0] == i && destination[1] == j) {
+            return true;
+        }
+
+        // already visited
+        if(maze[i][j] == -1) {
+            return false;
+        }
+
+        maze[i][j] = -1;
+
+        for(int[] dir : dirs) {
+
+            int r = i + dir[0];
+            int c = j + dir[1];
+
+            // roll until wall
+            while(r >= 0 && c >= 0 && r < m && c < n && maze[r][c] != 1) {
+                r += dir[0];
+                c += dir[1];
+            }
+
+            // move back to stopping point
+            r -= dir[0];
+            c -= dir[1];
+
+            if(dfs(maze, r, c, destination)) {
+                return true;
+            }
+        }
+
+        return false;
+    }
+}

town_judge.java

@@ -0,0 +1,32 @@
+// Time Complexity : O(n + m) where n is the number of people and m is the number of trust relationships
+// Space Complexity : O(n)
+// 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 in three sentences only
+// We use an indegree array to track the net trust score of every person.
+// If a person trusts someone, their score decreases; if they are trusted by someone, their score increases.
+// The town judge will have a score of n-1 because everyone trusts them and they trust nobody.
+
+class Solution {
+    public int findJudge(int n, int[][] trust) {
+
+        int[] indegrees = new int[n + 1];
+
+        // build trust scores
+        for(int[] t : trust) {
+            indegrees[t[0]]--;
+            indegrees[t[1]]++;
+        }
+
+        // judge should have score n-1
+        for(int i = 1; i <= n; i++) {
+            if(indegrees[i] == n - 1) {
+                return i;
+            }
+        }
+
+        return -1;
+    }
+}