From 63702cae6f1266eaca26591b27c6f2858b5352dd Mon Sep 17 00:00:00 2001
From: yashhh-23 <yashdedhia05@gmail.com>
Date: Fri, 31 Jul 2026 19:52:44 +0530
Subject: [PATCH] completed graph 1

---
 findjugde.java | 27 +++++++++++++++++++++++++++
 maze.java      | 43 +++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 70 insertions(+)
 create mode 100644 findjugde.java
 create mode 100644 maze.java

diff --git a/findjugde.java b/findjugde.java
new file mode 100644
index 0000000..19fc599
--- /dev/null
+++ b/findjugde.java
@@ -0,0 +1,27 @@
+// Time Complexity : O(n+e) where n is the number of nodes and e is the number of edges
+// Space Complexity : O(n) for storing the inDegree and outDegree arrays
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : having the thought of indegree and outdegree of the nodes
+
+// Your code here along with comments explaining your approach: started with the idea of indegree and outdegree of the nodes. The judge will have indegree of n-1 and outdegree of 0. So we can keep track of indegree and outdegree of each node and return the node which has indegree of n-1 and outdegree of 0.
+class Solution {
+    public int findJudge(int n, int[][] trust) {
+        int []inDegree=new int[n+1];
+        int []outDegree=new int[n+1];
+        for(int[]relation:trust)
+        {
+            int a=relation[0];
+            int b= relation[1];
+            outDegree[a]++;
+            inDegree[b]++;
+        }
+        for(int i=1;i<=n;i++)
+        {
+            if(outDegree[i]==0 && inDegree[i]==n-1)
+            {
+                    return i;
+            }
+        }
+        return -1;
+    }
+}
\ No newline at end of file
diff --git a/maze.java b/maze.java
new file mode 100644
index 0000000..3ddc238
--- /dev/null
+++ b/maze.java
@@ -0,0 +1,43 @@
+// Time Complexity : O(n*m) where n is the number of rows and m is the number of columns
+// Space Complexity : O(n*m) for the queue and the modified maze array
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : understanding the ball rolling mechanism and ensuring the ball stops at the destination
+// Your code here along with comments explaining your approach: used BFS to explore the maze. The ball can roll in four directions until it hits a wall. For each direction, I keep rolling the ball until it can't go further, then check if it has reached the destination. If not, I add the new position to the queue for further exploration. I also mark visited positions in the maze to avoid cycles.
+class Solution {
+    int[][] dirs;
+    int m,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;
+
+        Queue<int[]> q = new LinkedList<>();
+        q.add(new int[]{start[0], start[1]});
+        maze[start[0]][start[1]] = -1;
+
+        while(!q.isEmpty()){
+            int[] curr = q.poll();
+            for(int[] dir: dirs){
+                int r = dir[0] + curr[0];
+                int c = dir[1] + curr[1];
+
+                while(r>=0 && c>=0 && r<m && c<n && maze[r][c] != 1){
+                    r += dir[0];
+                    c += dir[1];
+                }
+
+                r -= dir[0];
+                c -= dir[1];
+
+                if(r == destination[0] && c == destination[1]) return true;
+                if(maze[r][c] != -1){
+                    q.add(new int[]{r,c});
+                    maze[r][c] = -1;
+                }
+            }
+        }
+
+        return false;
+    }
+}