Adding all the solutions

EmployeeImportance.java

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+// Time Complexity : O(n)
+// Spcae Complexity: O(n)
+// Explanaytion : I first store every employee in a HashMap using their id so that I can retrieve any employee in O(1) time. Then I use a queue to perform a BFS starting from the given id, adding all subordinates level by level. For each employee I dequeue, I add their importance to the total and push their subordinates into the queue.
+class Solution {
+    public int getImportance(List<Employee> employees, int id) {
+        Map<Integer, Employee> map = new HashMap<>();
+        Queue<Integer> q = new LinkedList<>();
+        for(int i=0; i<employees.size(); i++) {
+            map.put(employees.get(i).id, employees.get(i));
+        }
+        q.add(id);
+        int sum =0;
+        while(!q.isEmpty()) {
+            int currId = q.poll();
+            Employee curr = map.get(currId);
+            sum = sum + curr.importance;
+            for(int i=0; i <curr.subordinates.size(); i++) {
+                q.add(curr.subordinates.get(i));
+            }
+        }
+        return sum;
+    }
+}

RottingOranges.java

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+// TimeComplexity: O(m*n)
+// SpaceComplexity: O(m*n)
+// Explanation : I first traverse the grid to count all fresh oranges and add the positions of rotten oranges to a queue. Then I perform a BFS level by level, where each level represents one minute of time.
+//  For every rotten orange, I try to rot its four neighboring cells using the direction array. Whenever a fresh orange becomes rotten, 
+// I add it to the queue and decrease the fresh count. After BFS finishes, if no fresh oranges remain, I return the total time taken; otherwise, I return -1.
+
+class Solution {
+    public int orangesRotting(int[][] grid) {
+        int m = grid.length;
+        int n = grid[0].length;
+        int time =0;
+        int oneCount = 0;
+        Queue<int[]> q = new LinkedList<>();
+        for(int i =0; i<m; i++) {
+            for(int j=0; j<n; j++) {
+                if(grid[i][j] == 2) {
+                    q.add(new int[]{i,j});
+                }
+                if (grid[i][j] == 1) {
+                    oneCount++;
+                }
+            }
+        }
+        if(oneCount ==0) {
+            return 0;
+        }
+        int[][] dirs = {{0,-1}, {0,1}, {1,0}, {-1,0}};
+        while(!q.isEmpty()) {
+            int size = q.size();
+            for(int k=0; k<size; k++) {
+                int[] curr = q.poll();
+                for(int l =0 ; l<dirs.length; l++) {
+                    int i = dirs[l][0]+curr[0];
+                    int j = dirs[l][1]+curr[1];
+                    if(i>=0 && j>=0 && i<m && j<n) {
+                        if(grid[i][j] == 1){
+                            grid[i][j] =2;
+                            q.add(new int[]{i,j});
+                            oneCount--;
+                        }
+                    }
+
+                }
+
+            }
+            time++;
+        }
+        if(oneCount ==0) {
+            return time-1;
+        } 
+
+        return -1;
+
+    }
+}