diff --git a/DesignHashMap Linear Chaining Question b/DesignHashMap Linear Chaining Question
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+++ b/DesignHashMap Linear Chaining Question	
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+706. Design HashMap
+
+Design a HashMap without using any built-in hash table libraries.
+
+Implement the MyHashMap class:
+
+MyHashMap() initializes the object with an empty map.
+void put(int key, int value) inserts a (key, value) pair into the HashMap. If the key already exists in the map, update the corresponding value.
+int get(int key) returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
+void remove(key) removes the key and its corresponding value if the map contains the mapping for the key.
+
+
+Example 1:
+
+Input
+["MyHashMap", "put", "put", "get", "get", "put", "get", "remove", "get"]
+[[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
+Output
+[null, null, null, 1, -1, null, 1, null, -1]
+
+Explanation
+MyHashMap myHashMap = new MyHashMap();
+myHashMap.put(1, 1); // The map is now [[1,1]]
+myHashMap.put(2, 2); // The map is now [[1,1], [2,2]]
+myHashMap.get(1);    // return 1, The map is now [[1,1], [2,2]]
+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)
+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]]
+myHashMap.get(2);    // return -1 (i.e., not found), The map is now [[1,1]]
+
+
+Constraints:
+
+0 <= key, value <= 106
+At most 104 calls will be made to put, get, and remove.
\ No newline at end of file
diff --git a/HashMapDoubleHashing.java b/HashMapDoubleHashing.java
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+++ b/HashMapDoubleHashing.java
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+/*
+HashSet using Double Hashing is already done in Design 1 Assignment
+
+Time Complexity: O(1) for put, get, and remove.
+Space Complexity: O(1) for operations; O(n) for storage due to the 2D array.
+ */
+public class HashMapDoubleHashing {
+
+    int primaryArraySize;
+    int secondaryArraySize;
+    int[][] storage;
+
+    public HashMapDoubleHashing() {
+        this.primaryArraySize = 1001;
+        this.secondaryArraySize = 1000;
+        // Using -1 to indicate an empty slot since key and value are non-negative integers
+        this.storage = new int[primaryArraySize][];
+    }
+
+    private int getPrimaryHashValue(int key) {
+        return key / primaryArraySize;
+    }
+
+    private int getSecondaryHashValue(int key) {
+        return key % secondaryArraySize;
+    }
+
+    public void put(int key, int value) {
+        int primaryArrayIndex = getPrimaryHashValue(key);
+
+        if (storage[primaryArrayIndex] == null) {
+            // Initialize a new array if it doesn't exist, using -1 as the default value
+            this.storage[primaryArrayIndex] = new int[secondaryArraySize];
+            for (int i = 0; i < secondaryArraySize; i++) {
+                this.storage[primaryArrayIndex][i] = -1;
+            }
+        }
+
+        int secondaryArrayIndex = getSecondaryHashValue(key);
+        storage[primaryArrayIndex][secondaryArrayIndex] = value;
+    }
+
+    public int get(int key) {
+        int primaryArrayIndex = getPrimaryHashValue(key);
+
+        if (storage[primaryArrayIndex] == null) {
+            return -1; // Key doesn't exist
+        }
+
+        int secondaryArrayIndex = getSecondaryHashValue(key);
+        return storage[primaryArrayIndex][secondaryArrayIndex];
+    }
+
+    public void remove(int key) {
+        int primaryArrayIndex = getPrimaryHashValue(key);
+
+        if (storage[primaryArrayIndex] == null) {
+            return; // Nothing to remove as value doesn't exist
+        }
+
+        int secondaryArrayIndex = getSecondaryHashValue(key);
+        storage[primaryArrayIndex][secondaryArrayIndex] = -1; // Set to -1 to indicate removal
+    }
+}
diff --git a/HashMapLinearChainingHW.java b/HashMapLinearChainingHW.java
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+++ b/HashMapLinearChainingHW.java
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+/*
+    Time Complexity : {Time Complexity is analyzed based on => Whenever we are receiving some input, with the increase in our input, how our
+                        algorithm performs}
+                      In order to  optimize performance we are increasing the primary data structure size and decreasing the secondary data
+                      structure size to have fewer collisions. If our time complexity is not increasing after a particular point,and that point
+                      is a very small number, we can tell our average time complexity is O(1), here the small number for us is 100 in
+                      secondary data structure.
+
+    Note: If interviewer is asking that 100 size of secondary bucket is also large number, and it is affecting performance, then the answer can
+    be I can check in production and based on performance if time complexity is not O(1) or more, I can increase the size of primary data structure.
+
+    Note: It is also means increase of size of primary data structure and decrease of secondary data structure is trade off between
+    performance and space.
+
+    Space Complexity : For user oriented functions, all the functions put(), get(), remove() will be having amortized complexity O(1). For constructor,
+    we are creating storage so it will have O(n) but that is not what we be looking at. We only look at what we created for user.
+
+    Did this code successfully run on Leetcode : Yes
+
+    HASH MAP USING LINEAR CHAINING {HOMEWORK}
+
+    Here the primary dataset will be a Node Array and secondary will be linkedlist, in java hash map by default uses Linear chaining to
+    handle collisions
+
+    null, __ , null, null {primary buckets}
+           ↓
+        -1, -1            {linked list}
+           ↓
+        2001, 3
+           ↓
+        3001, 5
+
+    Here, like in previous homework Design 1, we choose 1000 for primary bucket and 1000 for secondary bucket, we followed this formula
+    where we took square root of 10^6 so we got 1000 Primary bucket and 1000 secondary bucket, but here if we use that size, we will have 1000
+    collisions.
+ */
+class HashMapLinearChainingHW {
+
+    class Node
+    {
+        int key;
+        int value;
+        Node next;
+
+        public Node(int key, int value) {
+            this.key = key;
+            this.value = value;
+        }
+    }
+
+    Node[] storage; //Primary array
+    int primaryBucketSize;
+
+    public HashMapLinearChainingHW() {
+        /*
+        As the bucket size is 10000, there will be 100 collisions, due to less collisions, after 100 elements, the time to access
+        the element, basically search will become constant i.e. O(1)
+        */
+        this.primaryBucketSize = 10000;
+        this.storage = new Node[primaryBucketSize];
+    }
+
+    private int getPrimaryHash(int key) {
+        return key%primaryBucketSize;
+    }
+
+    private Node getPrevNode(Node head, int key) {
+        Node prev = null;
+        Node curr = head;
+
+        //Traversing through the nodes to find previous
+        while(curr != null && curr.key!=key)
+        {
+            prev = curr;
+            curr = curr.next;
+        }
+
+        return prev;
+    }
+
+    public void put(int key, int value) {
+        int primaryIndex = getPrimaryHash(key);
+        if(storage[primaryIndex] ==  null){
+            storage[primaryIndex] = new Node(-1, -1);
+        }
+        /*
+        There will be two scenarios:
+        1) If the node exists, replace the value.
+        2) If he node doesn't exist, we will reach the end of linked list and the last node will point to null, so point the last node
+        to point to new node and new node will point to null.
+
+        Note: We will not need a tail pointer, as anyway due to point 1, we have to traverse through complete linked list to check if the
+        value exists, it doesn't then only we will add a new node in end, so tail pointer is of no use.
+         */
+
+        Node prev = getPrevNode(storage[primaryIndex], key);
+        if(prev.next == null) //Point 2
+        {
+            prev.next = new Node(key, value);
+        }
+        else {
+            prev.next.value = value; //Point 1
+        }
+    }
+
+    public int get(int key) {
+
+        int primaryIndex = getPrimaryHash(key);
+        if(storage[primaryIndex] ==  null) { //Primary index location in array itself is null
+            return -1;
+        }
+
+        Node prev = getPrevNode(storage[primaryIndex], key);
+        if(prev.next == null) { //Point 2, element not found after traversal
+            return -1;
+        }
+
+        return prev.next.value;
+    }
+
+    public void remove(int key) {
+        int primaryIndex = getPrimaryHash(key);
+        if(storage[primaryIndex] ==  null) { //Primary index location in array itself is null
+            return;
+        }
+
+        Node prev = getPrevNode(storage[primaryIndex], key);
+        if(prev.next == null) { //Point 2, element not found after traversal
+            return;
+        }
+
+        /*
+        eg: remove 3 hashmap {100,2 ; 101,3 ; 102,5}
+            prev = {100}
+            prev.next = curr.next => old value of prev.next is 101, new value is 102
+            curr.next = null meaning 101 is made to point to null
+        */
+        Node curr = prev.next;
+        prev.next = curr.next;
+        curr.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);
+ */
\ No newline at end of file
diff --git a/HashSetLinearChaining.java b/HashSetLinearChaining.java
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+++ b/HashSetLinearChaining.java
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+/*
+    Time Complexity : This implementation maintains the O(1) average time complexity for operations
+                       due to the hashing, similar to hash map.
+    Space Complexity : Amortized Complexity for User Functions (add(), remove(), contains()): O(1)
+                       Worst-Case Complexity for User Functions: O(m), where m is the number of elements in the set.
+ */
+class HashSetLinearChaining {
+
+    class Node {
+        int key;
+        Node next;
+
+        public Node(int key) {
+            this.key = key;
+        }
+    }
+
+    Node[] storage; // Primary array
+    int primaryBucketSize;
+
+    public HashSetLinearChaining() {
+        // Initialize with the same primary bucket size as the hash map
+        this.primaryBucketSize = 10000;
+        this.storage = new Node[primaryBucketSize];
+    }
+
+    private int getPrimaryHash(int key) {
+        return key % primaryBucketSize;
+    }
+
+    private Node getPrevNode(Node head, int key) {
+        Node prev = null;
+        Node curr = head;
+
+        // Traverse the nodes to find the previous node
+        while (curr != null && curr.key != key) {
+            prev = curr;
+            curr = curr.next;
+        }
+
+        return prev;
+    }
+
+    public void add(int key) {
+        int primaryIndex = getPrimaryHash(key);
+        if (storage[primaryIndex] == null) {
+            // Initialize the linked list with a dummy head node
+            storage[primaryIndex] = new Node(-1);
+        }
+
+        Node prev = getPrevNode(storage[primaryIndex], key);
+
+        // If the key does not exist, add it to the end of the list
+        if (prev.next == null) {
+            prev.next = new Node(key);
+        }
+    }
+
+    public void remove(int key) {
+        int primaryIndex = getPrimaryHash(key);
+        if (storage[primaryIndex] == null) {
+            // If the bucket is empty, do nothing
+            return;
+        }
+
+        Node prev = getPrevNode(storage[primaryIndex], key);
+        if (prev.next == null) {
+            // Key not found in the set
+            return;
+        }
+
+        // Remove the node
+        Node curr = prev.next;
+        prev.next = curr.next;
+        curr.next = null;
+    }
+
+    public boolean contains(int key) {
+        int primaryIndex = getPrimaryHash(key);
+        if (storage[primaryIndex] == null) {
+            // If the bucket is empty, the key is not present
+            return false;
+        }
+
+        Node prev = getPrevNode(storage[primaryIndex], key);
+        return prev.next != null; // Return true if the key is found
+    }
+}
+
+
diff --git a/QueueUsingStack Question b/QueueUsingStack Question
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+++ b/QueueUsingStack Question	
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+232. Implement Queue using Stacks {Easy}
+
+Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).
+
+Implement the MyQueue class:
+
+void push(int x) Pushes element x to the back of the queue.
+int pop() Removes the element from the front of the queue and returns it.
+int peek() Returns the element at the front of the queue.
+boolean empty() Returns true if the queue is empty, false otherwise.
+Notes:
+
+You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
+Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.
+
+
+Example 1:
+
+Input
+["MyQueue", "push", "push", "peek", "pop", "empty"]
+[[], [1], [2], [], [], []]
+Output
+[null, null, null, 1, 1, false]
+
+Explanation
+MyQueue myQueue = new MyQueue();
+myQueue.push(1); // queue is: [1]
+myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
+myQueue.peek(); // return 1
+myQueue.pop(); // return 1, queue is [2]
+myQueue.empty(); // return false
+
+
+Constraints:
+
+1 <= x <= 9
+At most 100 calls will be made to push, pop, peek, and empty.
+All the calls to pop and peek are valid.
+
+
+Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.
+
diff --git a/QueueUsingStacks.java b/QueueUsingStacks.java
new file mode 100644
index 00000000..2be333a0
--- /dev/null
+++ b/QueueUsingStacks.java
@@ -0,0 +1,140 @@
+/*
+ Time Complexity : For user oriented functions, push, pop, peep, empty time complexity is O(1). For pop, only once in a blue moon we will have
+                time complexity as O(n) because we have to move all elements to out stack from in stack once out stack is empty. But since it
+                is so rare, the average time complexity will be considered which is O(1) for selecting data structures.
+
+ Space Complexity : The space complexity of queue implementation with two stacks remains O(n) because the storage needed grows linearly with
+                    the number of elements in the queue.
+
+ Did this code successfully run on Leetcode : Yes
+
+Property of Queue: FIFO
+1) Two stacks will be used to implement queue , as asked in the problem statement
+9,5,6,7,2
+
+ |   2   |          |       |
+ |   7   |          |       |
+ |   6   |          |       |
+ |   5   |          |       |
+ |   9   |          |       |
+
+  IN STACK           OUT STACK
+
+2) POP,8,POP,10
+
+Now, we need to transfer all elements in Out stack before Pop to get the First element
+
+ |      |          |   9   |
+ |      |          |   5   |
+ |      |          |   6   |
+ |      |          |   7   |
+ |      |          |   2   |
+
+ POP
+
+ |      |          |       |
+ |      |          |   5   |
+ |      |          |   6   |
+ |      |          |   7   |
+ |      |          |   2   |
+
+ Insert 8
+ |      |          |       |
+ |      |          |   5   |
+ |      |          |   6   |
+ |      |          |   7   |
+ |   8  |          |   2   |
+
+  POP
+
+ |      |          |       |
+ |      |          |       |
+ |      |          |   6   |
+ |      |          |   7   |
+ |   8  |          |   2   |
+
+ Insert 10
+ |      |          |       |
+ |      |          |       |
+ |      |          |   6   |
+ |  10  |          |   7   |
+ |   8  |          |   2   |
+
+ 3) POP, POP, POP
+ |      |          |       |
+ |      |          |       |
+ |      |          |       |
+ |  10  |          |       |
+ |  8   |          |       |
+
+4) When Out array is empty transfer elements from In to Out
+ |      |          |       |
+ |      |          |       |
+ |      |          |       |
+ |      |          |   8   |
+ |      |          |   10  |
+
+ 5) Insert 11
+ |      |          |       |
+ |      |          |       |
+ |      |          |       |
+ |      |          |   8   |
+ |  11  |          |   10  |
+
+ */
+
+import java.util.Stack;
+
+class QueueUsingStacks {
+
+    Stack<Integer> inStack;
+    Stack<Integer> outStack;
+
+    public QueueUsingStacks() {
+        this.inStack = new Stack<>();
+        this.outStack = new Stack<>();
+    }
+
+    public void push(int x) {
+        inStack.push(x);
+    }
+
+    public int pop() {
+
+        /*
+        Edge case we don't need, but still writing it => we can't pop on an empty stack but in problem statement
+        it is mentioned All the calls to pop and peek are valid.
+        */
+        if(empty())
+            return -1;
+
+        peek();
+
+        return outStack.pop();
+    }
+
+    public int peek() {
+        if(outStack.isEmpty()){
+
+            while(!inStack.isEmpty()) {
+                outStack.push(inStack.pop());
+            }
+        }
+
+        return outStack.peek();
+    }
+
+    public boolean empty() {
+        //If both the stacks are empty then queue is 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();
+ */
\ No newline at end of file