From de14e712ded098567ae008ee15a6015d21fa5769 Mon Sep 17 00:00:00 2001
From: Sarvani Baru <sarvanibaru@gmail.com>
Date: Sat, 4 Apr 2026 17:25:31 -0700
Subject: [PATCH] Done Graph-1

---
 FindJudge.java | 28 ++++++++++++++++++++++++++
 TheMaze.java   | 53 ++++++++++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 81 insertions(+)
 create mode 100644 FindJudge.java
 create mode 100644 TheMaze.java

diff --git a/FindJudge.java b/FindJudge.java
new file mode 100644
index 0000000..679a887
--- /dev/null
+++ b/FindJudge.java
@@ -0,0 +1,28 @@
+// Time Complexity : O(E), number of trust relationships (edges)
+// Space Complexity : O(N) , number of people
+// 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
+/*
+We maintain a scores array to keep track of relationship scores for each person. We iterate through the trust
+relationships which is array of edges and decrement the score for person who is trusting i.e,x and increment
+the score for person who is getting trusted, i.e., y. Lastly, we go through the scores array and check if any
+person is trusted by all n-1 people, if so, we return that person's label. If not, -1.
+ */
+class Solution {
+    public int findJudge(int n, int[][] trust) {
+        int[] scores = new int[n + 1];
+
+        for(int[] relation : trust) {
+            scores[relation[0]]--;
+            scores[relation[1]]++;
+        }
+
+        for(int i = 1 ; i <= n ; i++) {
+            if(scores[i] == n - 1)
+                return i;
+        }
+        return -1;
+    }
+}
\ No newline at end of file
diff --git a/TheMaze.java b/TheMaze.java
new file mode 100644
index 0000000..9c30b71
--- /dev/null
+++ b/TheMaze.java
@@ -0,0 +1,53 @@
+// Time Complexity : O(mn)
+// Space Complexity : O(mn)
+// 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
+/*
+We do a dfs recursive approach to roll the ball from the start position until destination coordinates are met.
+As mentioned in the problem, the ball can be rolled in 4 directions, so we declare a dirs array and iterate the
+i,j coordinates in all 4 directions. We need to roll ball until a wall is hit, so we use a while loop to iterate
+with the given direction until conditions go out of bounds.Now, we decrement to the previous position and explore
+in all directions from that position and we also keep track of visited array to mark positions such that we
+dont visit them again.
+ */
+class Solution {
+    boolean flag;
+    int[][] dirs;
+    int m, n;
+    public boolean hasPath(int[][] maze, int[] start, int[] destination) {
+        this.flag = false;
+        this.dirs = new int[][] {{-1, 0} , {0, -1} , {1, 0} , {0, 1}};
+        this.m = maze.length;
+        this.n = maze[0].length;
+        boolean[][] visited = new boolean[m][n];
+        dfs(maze, start[0] , start[1], destination, visited);
+        return flag;
+    }
+
+    private void dfs(int[][] maze, int i , int j, int[] destination, boolean[][] visited) {
+        if(i == destination[0] && j == destination[1]) {
+            flag = true;
+            return;
+        }
+
+        visited[i][j] = true;
+
+        for(int[] dir : dirs) {
+            int nr = dir[0] + i;
+            int nc = dir[1] + j;
+
+            while(nr >= 0 && nc >= 0 && nr < m && nc < n && maze[nr][nc] == 0) {
+                nr += dir[0];
+                nc += dir[1];
+            }
+
+            nr -= dir[0];
+            nc -= dir[1];
+
+            if(!visited[nr][nc])
+                dfs(maze, nr, nc , destination, visited);
+        }
+    }
+}
\ No newline at end of file