From 063beb0ce58cdb868ac2ffd77590446950a26870 Mon Sep 17 00:00:00 2001
From: candida17 <53035166+candida17@users.noreply.github.com>
Date: Fri, 17 Oct 2025 15:03:57 -0700
Subject: [PATCH 1/2] hasPath

---
 Problem2.java | 68 +++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 68 insertions(+)
 create mode 100644 Problem2.java

diff --git a/Problem2.java b/Problem2.java
new file mode 100644
index 0000000..c7b79bb
--- /dev/null
+++ b/Problem2.java
@@ -0,0 +1,68 @@
+// Time Complexity : O(m*n)
+// Space Complexity :O(m*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
+//Using BFS traversal we start by adding the start cell in the queue
+//we then traverse  up, down right and left of the maze until we find a wall and add them into the queue
+//To prevent the visit of same cell again we mark the cells as visited whenever its being added to the queue
+//If the destination is present inside the queue we return true
+
+import java.util.*;
+class Solution {
+    public boolean hasPath(int[][] maze, int[] start, int[] destination) {
+        int m = maze.length;
+        int n = maze[0].length;
+        int [][] dirs = new int[][] {{-1,0}, {1,0}, {0,-1}, {0,1}};
+        if(start[0] == destination[0] && start[1] == destination[1]) return true;
+        Queue<int []> q = new LinkedList<>();
+        q.add(start);//add the start indices into the queue (0.4)
+        maze[start[0]][start[1]] = 2; //mark the start indices as visited ie 0,4 is visited
+        
+        while(!q.isEmpty()) {
+            int[] cur = q.poll();
+            for(int[] dir: dirs) {
+                int r = cur[0];
+                int c = cur[1];
+                //the next 0 to go in the queue is not the immediate neighbour but the one above the obstacle or wall
+                while(r < m && r >=0 && c>=0 && c< n && maze[r][c] != 1) {
+                    r += dir[0];
+                    c += dir[1];
+                }
+                r -= dir[0];
+                c -= dir[1];
+                //check if r and c are the destination
+                if(r== destination[0] && c == destination[1]) return true;
+                //only if the neighbours are not marked visited add them to queue
+                if(maze[r][c] !=2) {
+                    maze[r][c]= 2;
+                    q.add(new int[]{r,c});
+                }
+            }
+            
+        }
+        return false;
+        
+    }
+}
+public class Main {
+    public static void main(String[] args) {
+        Solution sol = new Solution();
+
+        int[][] maze = {
+            {0, 0, 1, 0, 0},
+            {0, 0, 0, 0, 0},
+            {0, 0, 0, 1, 0},
+            {1, 1, 0, 1, 1},
+            {0, 0, 0, 0, 0}
+        };
+
+        int[] start = {0, 4};
+        int[] destination = {4, 4};
+
+        boolean result = sol.hasPath(maze, start, destination);
+        System.out.println(result);
+    }
+}

From bf73795ab801e1b011852d2a8b7a1eda2bdb02a0 Mon Sep 17 00:00:00 2001
From: candida17 <53035166+candida17@users.noreply.github.com>
Date: Fri, 17 Oct 2025 15:28:53 -0700
Subject: [PATCH 2/2] findJudge

---
 Problem1.java | 26 ++++++++++++++++++++++++++
 1 file changed, 26 insertions(+)
 create mode 100644 Problem1.java

diff --git a/Problem1.java b/Problem1.java
new file mode 100644
index 0000000..91eef70
--- /dev/null
+++ b/Problem1.java
@@ -0,0 +1,26 @@
+// Time Complexity : O(V + E), where V is number of people and E is number of trust pairs.
+// Space Complexity :O(V), for storing indegrees and outDegrees arrays.
+// 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 keep track of number of people trusting each other using indegree and outdegree
+//when a trusts b outdeg will decrease and b indeg will increase
+//in the end we need to check if the outdeg becomes 0 to satisfy the first condition - The town judge trusts nobody.
+//and check for second condition Everybody (except for the town judge) trusts the town judge. - check if indeg becomes equal to 1 less number of people
+class Solution {
+    public int findJudge(int n, int[][] trust) {
+        int[] outdeg = new int[n+1];
+        int[] indeg = new int[n+1];
+        for (int[] tr: trust) {
+            outdeg[tr[0]]--;
+            indeg[tr[1]]++;
+        }
+
+        for(int i = 1; i <= n; i++) {
+            if(outdeg[i] == 0 && indeg[i] == n-1) return i;
+        }
+        return -1;
+    }
+}