diff --git a/MinStack.java b/MinStack.java
new file mode 100644
index 00000000..87a6616d
--- /dev/null
+++ b/MinStack.java
@@ -0,0 +1,55 @@
+// Time Complexity : O(1) for all operations
+// Space Complexity : O(n) where n is the number of elements in the stack
+// Did this code successfully run on Leetcode : Yes
+// Approach: I am using 2 stacks. One stack is used to store the elements and the other stack is used to store the minimum element at that point. I am pushing the element to the stack and also pushing the minimum element to the minStack. If the element is removed from the stack, I am removing the element from the stack and also removing the element from the minStack. If the minStack is empty, I am setting the min value to Integer.MAX_VALUE. If the minStack is not empty, I am setting the min value to the top element of the minStack.
+
+import java.util.Stack;
+
+class MinStack {
+
+    Stack<Integer> stack;
+    Stack<Integer> minStack;
+    int min = Integer.MAX_VALUE;
+
+    public MinStack() {
+        stack = new Stack<>();
+        minStack = new Stack<>();
+    }
+
+    public void push(int val) {
+        min = Math.min(val, min);
+        stack.push(val);
+        minStack.push(min);
+    }
+
+    public void pop() {
+        if(!stack.isEmpty()){
+            stack.pop();
+        }
+        if(!minStack.isEmpty()){
+            minStack.pop();
+        }
+        if(!minStack.isEmpty()){
+            min = minStack.peek();
+        }else{
+            min = Integer.MAX_VALUE;
+        }
+    }
+
+    public int top() {
+        return stack.peek();
+    }
+
+    public int getMin() {
+        return min;
+    }
+}
+
+/**
+ * Your MinStack object will be instantiated and called as such:
+ * MinStack obj = new MinStack();
+ * obj.push(val);
+ * obj.pop();
+ * int param_3 = obj.top();
+ * int param_4 = obj.getMin();
+ */
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diff --git a/MyHashSet.java b/MyHashSet.java
new file mode 100644
index 00000000..34175cd4
--- /dev/null
+++ b/MyHashSet.java
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+// Time Complexity : O(1) for all operations
+// Space Complexity : O(n) where n is the number of elements in the hashset
+// Did this code successfully run on Leetcode : Yes
+// Approach: I am using 2D array where primary array is of size 1000 and secondary array is of size 1000. I am using 2 hash functions to calculate the index of the primary and secondary array. I am using boolean values to store the values in the hashset. If the value is present in the hashset, I am setting the value to true. If the value is removed from the hashset, I am setting the value to false. If the value is not present in the hashset, I am returning false.
+
+class MyHashSet {
+
+    boolean[][] values;
+    int buckets;
+    int bucketItems;
+
+    public MyHashSet() {
+        buckets = 1000;
+        bucketItems = 1000;
+        values = new boolean[buckets][];
+    }
+
+    private int hash1(int key){
+        return key%buckets;
+    }
+
+    private int hash2(int key){
+        return key/bucketItems;
+    }
+
+    public void add(int key) {
+        int index1 = hash1(key);
+        int index2 = hash2(key);
+
+        if(values[index1] == null){
+            if(index1 == 0){
+                values[index1] = new boolean[bucketItems+1];
+            }else{
+                values[index1] = new boolean[bucketItems];
+            }
+        }
+        values[index1][index2] = true;
+    }
+
+    public void remove(int key) {
+        int index1 = hash1(key);
+        int index2 = hash2(key);
+
+        if(values[index1] == null){
+            return;
+        }else{
+            values[index1][index2] = false;
+        }
+    }
+
+    public boolean contains(int key) {
+        int index1 = hash1(key);
+        int index2 = hash2(key);
+
+        if(values[index1] == null){
+            return false;
+        }
+        return values[index1][index2];
+    }
+}
+
+/**
+ * Your MyHashSet object will be instantiated and called as such:
+ * MyHashSet obj = new MyHashSet();
+ * obj.add(key);
+ * obj.remove(key);
+ * boolean param_3 = obj.contains(key);
+ */
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