Completed Design-2

DesignHashMap.java

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+// Time Complexity : O(1)
+// Space Complexity : O(n) where n is number of key value pairs (O(n+k), but k is 1000 buckets - O(1))
+// 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
+/*
+maintain an array of linked list.For any inout, get the hash to check if it has previous occurence or not.
+If it has, fetch the corresponding previous node to put/remove/access the next node. If not, create a new
+node along with creation of dummy node first for easier access.
+ */
+
+class MyHashMap {
+    Node[] storage;
+    int buckets;
+
+    class Node {
+        int key;
+        int value;
+        Node next;
+
+        public Node(int key, int value) {
+            this.key = key;
+            this.value = value;
+        }
+    }
+
+    public MyHashMap() {
+        this.buckets = 1000;
+        this.storage = new Node[buckets];
+    }
+
+    private int getHash(int key) {
+        return key % buckets;
+    }
+
+    private Node getPrev(Node head, int key) {
+        Node prev = null;
+        Node curr = head;
+        while(curr != null && curr.key != key) {
+            prev = curr;
+            curr = curr.next;
+        }
+        return prev;
+    }
+
+    public void put(int key, int value) {
+        int index = getHash(key);
+        if(storage[index] == null) {
+            storage[index] = new Node(-1, -1);
+            storage[index].next = new Node(key, value);
+            return;
+        }
+        Node prev = getPrev(storage[index], key);
+        if(prev.next == null)
+            prev.next = new Node(key, value);
+        else
+            prev.next.value = value;
+    }
+
+    public int get(int key) {
+        int index = getHash(key);
+        if(storage[index] == null)
+            return -1;
+        Node prev = getPrev(storage[index], key);
+        if(prev.next == null)
+            return -1;
+        return prev.next.value;
+    }
+
+    public void remove(int key) {
+        int index = getHash(key);
+        if(storage[index] == null) {
+            return;
+        }
+        Node prev = getPrev(storage[index] , key);
+        if(prev.next == null)
+            return;
+        prev.next = prev.next.next;
+
+    }
+}
+
+/**
+ * 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);
+ */

ImplementQueueusingStacks.java

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+// Time Complexity : Amortized O(1) average case, worst case - O(n) for pop & peek
+// Space Complexity : O(1)
+// 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
+/*
+Maintain 2 stacks one for push operation and other for pop, peek operations. Whenever pop/peek happens,
+push all the elements from inStack to outStack if outStack is empty, this way, we can have queue property
+satisfied whenever peek or pop are called. As per question, all calls to pop and peek are valid, so
+we need not check explicitly, if not, we need to validate inStack is not empty in those methods and return
+-1 if so.
+ */
+
+class MyQueue {
+    Stack<Integer> inStack;
+    Stack<Integer> outStack;
+
+    public MyQueue() {
+        inStack = new Stack<>();
+        outStack = new Stack<>();
+    }
+
+    public void push(int x) {
+        inStack.push(x);
+    }
+
+    public int pop() {
+        peek();
+        return outStack.pop();
+    }
+
+    public int peek() {
+        if(outStack.isEmpty()) {
+            while(!inStack.isEmpty()) {
+                outStack.push(inStack.pop());
+            }
+        }
+        return outStack.peek();
+    }
+
+    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();
+ */