diff --git a/EmployeeImportance.java b/EmployeeImportance.java
new file mode 100644
index 0000000..038cf98
--- /dev/null
+++ b/EmployeeImportance.java
@@ -0,0 +1,42 @@
+/*
+// Definition for Employee.
+class Employee {
+    public int id;
+    public int importance;
+    public List<Integer> subordinates;
+};
+*/
+
+class Solution {
+    // TC - O(n)
+    // SC - O(n)
+    /*
+        This solution uses DFS to calculate the total importance of an employee and all of their subordinates.
+        A HashMap is first created to map each employee's ID to their Employee object, allowing O(1) lookups.
+        Starting from the given employee ID, the DFS adds the employee's own importance to the total.
+        It then recursively visits each subordinate and adds their importance as well.
+        This process continues until all direct and indirect subordinates have been visited.
+        The final result is the sum of the importance values of the employee and everyone under them.
+    
+     */
+    public int getImportance(List<Employee> employees, int id) {
+        HashMap<Integer, Employee> map = new HashMap<>();
+        for (Employee e : employees) {
+            map.put(e.id, e);
+        }
+
+        return dfs(map, id);
+    }
+
+    private int dfs(HashMap<Integer, Employee> map, int id) {
+        Employee e = map.get(id);
+        int result = 0;
+        result += e.importance;
+
+        for (int eid : e.subordinates) {
+            result += dfs(map, eid);
+        }
+
+        return result;
+    }
+}
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diff --git a/RottingOranges.java b/RottingOranges.java
new file mode 100644
index 0000000..61cb709
--- /dev/null
+++ b/RottingOranges.java
@@ -0,0 +1,59 @@
+class Solution {
+    /*
+        TC - O(m * n)
+        SC - O(m * n)
+        This solution uses multi-source BFS to simulate how the oranges rot minute by minute.
+        First, all initially rotten oranges are added to a queue, and the number of fresh oranges is counted.
+        If there are no fresh oranges, the answer is 0.
+        The BFS processes the queue level by level, where each level represents one minute.
+        For every rotten orange, its four neighboring cells are checked. If a neighboring orange is fresh, it becomes rotten, is added to the queue, and the fresh orange count is decreased.
+        After processing one level of the queue, one minute has passed, so time is incremented.
+        When the BFS finishes:
+        If any fresh oranges remain, they could not be reached, so return -1.
+        Otherwise, return time - 1, since the last increment happens after processing the final level.
+     */
+    public int orangesRotting(int[][] grid) {
+        Queue<int[]> q = new LinkedList<>();
+        int freshCount = 0;
+
+        for (int i = 0; i < grid.length; i++) {
+            for (int j = 0; j < grid[0].length; j++) {
+                if (grid[i][j] == 2) {
+                    q.add(new int[]{i, j});
+                }
+                if (grid[i][j] == 1) {
+                    freshCount++;
+                }
+            }
+        }
+
+        if (freshCount == 0) {
+            return 0;
+        }
+
+        int[][] dirs = new int[][]{{1, 0}, {0, 1}, {0, -1}, {-1, 0}};
+        int time = 0;
+
+        while (!q.isEmpty()) {
+            int size = q.size();
+            for (int i = 0; i < size; i++) {
+                int[] curr = q.poll();
+                for (int[] dir : dirs) {
+                    int nr = curr[0] + dir[0];
+                    int nc = curr[1] + dir[1];
+
+                    if (nr >= 0 && nc >= 0 && nr < grid.length && nc < grid[0].length && grid[nr][nc] == 1) {
+                        q.add(new int[]{nr, nc});
+                        grid[nr][nc] = 2;
+                        freshCount--;
+                    }
+                }
+            } 
+            time++;
+        }
+
+        if (freshCount != 0) return -1;
+
+        return time - 1;
+    }
+}
\ No newline at end of file