diff --git a/HashMap.java b/HashMap.java
new file mode 100644
index 00000000..8bc5c96b
--- /dev/null
+++ b/HashMap.java
@@ -0,0 +1,112 @@
+// ## Problem 2:Design Hashmap (https://leetcode.com/problems/design-hashmap/)
+
+// Time Complexity : Avg ~ O(1) | Worst ~ O(100) since the linear chaining will go at most 100 layers 
+// Space Complexity : O(n) ~ Storage array
+// 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
+/*
+ * Initialzed the storage array of ListNode to 10000
+ * Used dummy, prev and curr pointer to perform the operations
+ * When idx is not used, create dummy node and then check if ListNode of the key already exists. Same for remove.
+ */
+
+public class HashMap {
+
+    class ListNode {
+        int key;
+        int value;
+        ListNode next;
+
+        public ListNode(int key, int value) {
+            this.key = key;
+            this.value = value;
+        }
+    }
+
+    private int fixedSize = 10000;
+    ListNode[] storage;
+
+    public HashMap() {
+        storage = new ListNode[fixedSize];
+    }
+
+    private ListNode getPrevNode(ListNode dummy, int key) {
+        ListNode prev = dummy;
+        ListNode curr = prev.next;
+
+        while(curr != null && curr.key != key) {
+            prev = curr;
+            curr = curr.next;
+        }
+        return prev;
+    }
+
+    private int computeHash(int key) {
+        return key % fixedSize;
+    }
+
+    public void put(int key, int value) {
+        int idx = computeHash(key);
+
+        if(storage[idx] == null) {
+            ListNode dummy = new ListNode(-1, -1);
+            storage[idx] = dummy;
+        } 
+
+        ListNode prev = getPrevNode(storage[idx], key);
+
+        if(prev.next == null) {
+            ListNode newNode = new ListNode(key, value);
+            prev.next = newNode;
+        } else {
+            prev.next.value = value;
+        }
+    }
+    
+    public int get(int key) {
+        int idx = computeHash(key);
+        if(storage[idx] == null) {
+            return -1;
+        }
+
+        ListNode curr = storage[idx];
+        while(curr!=null) {
+            if(curr.key == key) {
+                return curr.value;
+            } else {
+                curr = curr.next;
+            }
+        }
+
+        return -1;
+    }
+    
+    public void remove(int key) {
+        int idx = computeHash(key);
+        if(storage[idx] == null) {
+            return;
+        }
+
+        ListNode prev = getPrevNode(storage[idx], key);
+    
+        if(prev.next != null) {
+            prev.next = prev.next.next;
+        }
+    }
+
+    public static void main(String[] args) {
+        HashMap myHashMap = new HashMap();
+        myHashMap.put(1, 1); // The map is now [[1,1]]
+        myHashMap.put(2, 2); // The map is now [[1,1], [2,2]]
+        System.out.println(myHashMap.get(1));    // return 1, The map is now [[1,1], [2,2]]
+        System.out.println(myHashMap.get(3));    // return -1 (i.e., not found), The map is now [[1,1], [2,2]]
+        myHashMap.put(2, 1); // The map is now [[1,1], [2,1]] (i.e., update the existing value)
+        System.out.println(myHashMap.get(2));   // return 1, The map is now [[1,1], [2,1]]
+        myHashMap.remove(2); // remove the mapping for 2, The map is now [[1,1]]
+        System.out.println(myHashMap.get(2));   // return -1 (i.e., not found), The map is now [[1,1]]
+
+    }
+}
diff --git a/QueueUsingStacks.java b/QueueUsingStacks.java
new file mode 100644
index 00000000..3e2541ae
--- /dev/null
+++ b/QueueUsingStacks.java
@@ -0,0 +1,61 @@
+// ## Problem 1: (https://leetcode.com/problems/implement-queue-using-stacks/)
+
+// Time Complexity : Avg ~ O(1) Worst ~ O(100) since the problem mentions atmost 100 calls will be made
+// Space Complexity : O(2n) ~ O(n) for 2 stacks
+// 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
+/*
+ * Took two stacks similar to what was explained in the class.
+ * When pop and peek operation is triggered and outStack is Empty, start transferring elements from inStack
+ * Always return pop and peek from outStack and push in inStack 
+ */
+
+import java.util.*;
+
+public class QueueUsingStacks {
+    Stack<Integer> inStack;
+    Stack<Integer> outStack;
+
+    public QueueUsingStacks() {
+        inStack = new Stack<>();
+        outStack = new Stack<>();
+    }
+
+    public void push(int x) {
+        inStack.push(x);
+    }
+    
+    public int pop() {
+        transferElementsFromStack();
+        return outStack.pop();
+    }
+    
+    public int peek() {
+        transferElementsFromStack();
+        return outStack.peek();
+    }
+    
+    public boolean empty() {
+        return inStack.isEmpty() && outStack.isEmpty();
+    }
+
+    private void transferElementsFromStack() {
+        if(outStack.isEmpty()) {
+            while(!inStack.isEmpty()) {
+                outStack.push(inStack.pop());
+            }
+        }
+    }
+
+    public static void main(String[] args) {
+        QueueUsingStacks myQueue = new QueueUsingStacks();
+        myQueue.push(1); // queue is: [1]
+        myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
+        System.out.println(myQueue.peek());// return 1
+        System.out.println(myQueue.pop()); // return 1, queue is [2]
+        System.out.println(myQueue.empty()); // return false
+    }
+}
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