diff --git a/EmployeeImportance.java b/EmployeeImportance.java
new file mode 100644
index 0000000..af850d4
--- /dev/null
+++ b/EmployeeImportance.java
@@ -0,0 +1,44 @@
+// Time Complexity : O(n)
+// 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
+/*
+Feed the combination of employee id and their respective object into a map for optimized search. Now, have
+a queue to start with the given input employee and then iteratively traverse all his subordinates by
+summing up the importance at each traversal and we can return it.
+ */
+/*
+// Definition for Employee.
+class Employee {
+    public int id;
+    public int importance;
+    public List<Integer> subordinates;
+};
+*/
+
+class Solution {
+    Map<Integer, Employee> map;
+    public int getImportance(List<Employee> employees, int id) {
+        this.map = new HashMap<>();
+        for(Employee e : employees) {
+            map.put(e.id, e);
+        }
+
+        Queue<Employee> q = new LinkedList<>();
+        q.add(map.get(id));
+        int totalImportance = 0;
+
+        while(!q.isEmpty()) {
+            Employee e = q.poll();
+            totalImportance += e.importance;
+            if(e.subordinates != null) {
+                for(int subordinate : e.subordinates) {
+                    q.add(map.get(subordinate));
+                }
+            }
+        }
+        return totalImportance;
+    }
+}
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diff --git a/RottingOranges.java b/RottingOranges.java
new file mode 100644
index 0000000..5a76ebc
--- /dev/null
+++ b/RottingOranges.java
@@ -0,0 +1,60 @@
+// 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
+/*
+The idea is to first identify the rotten orange positions and push them into queue.Now, traverse its
+nieghbours using a directions array and mark those fresh oranges(if present) as rotten and push them into
+queue again as they get the eligibility because of being rotten.We can keep track of fresh oranges at the
+start and at end of traversal to compare and see if all fresh oranges are rotten or not.We can leverage
+time variable and increment it at each level of traversal.
+ */
+class Solution {
+    int[][] dirs;
+    int m, n;
+    public int orangesRotting(int[][] grid) {
+        this.dirs = new int[][] {{-1,0}, {0, -1} , {1, 0}, {0, 1}};
+        int time = 0;
+        Queue<int[]> q = new LinkedList<>();
+        this.m = grid.length;
+        this.n = grid[0].length;
+        int fresh = 0;
+
+        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});
+                else if(grid[i][j] == 1)
+                    fresh++;
+            }
+        }
+
+        if(fresh == 0)
+            return time;
+
+        while(!q.isEmpty()) {
+            int size = q.size();
+            time++;
+            for(int i = 0 ; i < size ; i++) {
+                int[] curr = q.poll();
+                for(int[] dir : dirs) {
+                    int r = curr[0] + dir[0];
+                    int c = curr[1] + dir[1];
+                    if(r >= 0 && c >= 0 && r < m && c < n && grid[r][c] == 1) {
+                        grid[r][c] = 2;
+                        q.add(new int[]{r, c});
+                        fresh--;
+
+                        if(fresh == 0)
+                            return time;
+                    }
+                }
+
+            }
+        }
+
+        return -1;
+    }
+}
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