diff --git a/MyHashMap.java b/MyHashMap.java
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--- /dev/null
+++ b/MyHashMap.java
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+// Time Complexity : O(1) for put(), get() and remove() operations.
+// Space Complexity : O(n) where n is the number of key-value pairs stored in the hashmap.
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+/*
+ * Problem: Design HashMap
+ * Approach: Using Arrays, LinkedList and Hashing - We use the hash function to locate the
+ * index of the bucket. We then use a find() function to retrieve the data for the key using the index from hashing.
+ * We use the ref to the prev node so that it works for all three cases: put, get and remove 
+ *
+*/
+
+class MyHashMap {
+    class Node {
+        int key;
+        int val;
+        Node next;
+
+        public Node(int key, int val) {
+            this.key = key;
+            this.val = val;
+        }
+    }
+
+    // We define the bucket size to be 10^4 considering the constraints.
+    private static final int BUCKET_SIZE = 10000;
+    // We use an array of linked list nodes for storing the key values.
+    private Node[] storage;
+    
+    public MyHashMap() {
+        this.storage = new Node[BUCKET_SIZE];
+    }
+
+    // returns the bucket index after hashing 
+    private int hashFunction(int key) {
+        return key % BUCKET_SIZE;
+    }
+
+    // finds the key within the provided linked list by iterating through all elements
+    private Node find(Node head, int key) {
+        // returns previous node to current node if we find the current key in the linked list
+        Node prev = head;
+        Node curr = head.next;
+
+        while (curr != null && curr.key != key) {
+            prev = curr;
+            curr = curr.next;
+        }
+
+        return prev;
+    }
+    
+    // puts the key-value pair in the map if not present and overwrites the value if key is already present
+    public void put(int key, int value) {
+        int idx = hashFunction(key);
+
+        // check if a linked list is present, if not initialize a linked list with a dummy node
+        if (storage[idx] == null) {
+            storage[idx] = new Node(-1, -1);
+        }
+
+        Node prev = find(storage[idx], key);
+
+        if (prev.next == null) {
+            prev.next = new Node(key, value);
+        } else {
+            prev.next.val = value;
+        }
+    }
+    
+    // gets the value for a key
+    public int get(int key) {
+        int idx = hashFunction(key);
+
+        // check if a linked list is present, if not return -1 as the key does not exist
+        if (storage[idx] == null) return -1;
+
+        Node prev = find(storage[idx], key);
+
+        if (prev.next == null) return -1;
+
+        return prev.next.val;
+    }
+    
+    // removes the linked list node corresponding to the specific input key
+    public void remove(int key) {
+        int idx = hashFunction(key);
+
+        // check if a linked list is present, if not return as there is nothing to remove
+        if (storage[idx] == null) return;
+
+        Node prev = find(storage[idx], key);
+
+        if (prev.next == null) return;
+
+        Node temp = prev.next;
+        prev.next = prev.next.next;
+        temp.next = null;
+    }
+}
+
+/**
+ * Your MyHashMap object will be instantiated and called as such:
+ * MyHashMap obj = new MyHashMap();
+ * obj.put(key,value);
+ * int param_2 = obj.get(key);
+ * obj.remove(key);
+ */
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diff --git a/MyQueue.java b/MyQueue.java
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+++ b/MyQueue.java
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+// Time Complexity : O(1) for push(), 
+//                   Amortized O(1) for pop() and O(n) in worst case, 
+//                   Amortized O(1) for peek() and O(n) in worst case, 
+//                   O(1) for empty()
+
+// Space Complexity : O(n) as we need auxiliary space to save elements.
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+/*
+ * Problem: Implement Queue using Stacks
+ * Approach: Two stacks
+ * We use an instack for pushing elements to the queue and an outstack 
+ * for popping elements off the queue as the elements in instack require re-arranging to respect the queue ordering.
+ * We do the same while peeking elements. We transfer the elements between the stacks as and when required.
+ *
+*/
+
+import java.util.Stack;
+
+class MyQueue {
+    Stack<Integer> inStack;
+    Stack<Integer> outStack;
+
+    public MyQueue() {
+        inStack = new Stack<>();
+        outStack = new Stack<>();
+    }
+    
+    // Pushes element to the back of the queue
+    public void push(int x) {
+        inStack.push(x);
+    }
+    
+    // Pops element from the front of the queue and returns it. We need to move all elements from instack to outstack if
+    // outstack is empty in order to maintain the elements in right order of first-in-first-out.
+    public int pop() {
+        if (outStack.isEmpty()) {
+            while (!inStack.isEmpty()) {
+                outStack.push(inStack.pop());
+            }
+        }
+
+        return outStack.pop();
+    }
+    
+    // Returns the element at the front of the queue
+    public int peek() {
+        if (outStack.isEmpty()) {
+            while (!inStack.isEmpty()) {
+                outStack.push(inStack.pop());
+            }
+        }
+        return outStack.peek();
+    }
+    
+    // Checks whether the queue is empty
+    public boolean empty() {
+        return inStack.isEmpty() && outStack.isEmpty();
+    }
+}
+
+/**
+ * Your MyQueue object will be instantiated and called as such:
+ * MyQueue obj = new MyQueue();
+ * obj.push(x);
+ * int param_2 = obj.pop();
+ * int param_3 = obj.peek();
+ * boolean param_4 = obj.empty();
+ */
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