Design-2 problems solution

MyHashSet.java

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+//Design a Hash Set - Double Hashing Solution
+//Time Complexity : O(1)
+//Space Complexity : O(n)
+
+public class MyHashSet {
+    boolean[][] set;
+    int pbucket;
+    int sbucket;
+
+    public MyHashSet() {
+        this.pbucket = 1000;
+        this.sbucket = 1000;
+        this.set = new boolean[pbucket][];
+    }
+
+    public void add(int key) {
+        int hash1 = key%pbucket;
+        if(set[hash1] == null) {
+            //initiate the array
+            if(hash1 == 0)
+                set[hash1] = new boolean[sbucket+1];
+            else
+                set[hash1] = new boolean[sbucket];
+        }
+        int hash2 = key/sbucket;
+        set[hash1][hash2] = true;
+    }
+
+    public void remove(int key) {
+        int hash1 = key%pbucket;
+        if(set[hash1] != null) {
+            int hash2 = key/sbucket;
+            set[hash1][hash2] = false;
+        }
+    }
+
+    public boolean contains(int key) {
+        int hash1 = key%pbucket;
+        if(set[hash1] == null) {
+            return false;
+        }
+        int hash2 = key/sbucket;
+        return set[hash1][hash2];
+    }
+}

MyQueue.java

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+//Implement Queue using Stacks
+//Time Complexity: O(1) for push, O(n) for pop and peek in worst case, O(1) for pop and peek Amortized
+//Space Complexity: O(n)
+
+import java.util.Stack;
+
+class MyQueue {
+
+    Stack<Integer> in;
+    Stack<Integer> out;
+
+    public MyQueue() {
+        in = new Stack<>();
+        out = new Stack<>();
+    }
+
+    public void push(int x) {
+        in.push(x);
+    }
+
+    public int pop() {
+        if(empty()) return -1;
+        peek();
+        return out.pop();
+    }
+
+    public int peek() {
+        if(out.isEmpty()) {
+            while(!in.isEmpty()) {
+                out.push(in.pop());
+            }
+        }
+        return out.peek();
+    }
+
+    public boolean empty() {
+        return in.isEmpty() && out.isEmpty();
+    }
+}