diff --git a/Problem1.java b/Problem1.java
new file mode 100644
index 00000000..f9fe683b
--- /dev/null
+++ b/Problem1.java
@@ -0,0 +1,70 @@
+// Time Complexity :O(1)
+// Space Complexity : O(n)
+// 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
+/* we are implementing HashSet which is 1D boolean array with primary and secondary hashing to store keys. 
+we have followed double hashing technique to avoid collisions.
+add, remove, and contains operations ad implemented with O(1) time complexity.
+time - O(1)
+space - O(n)*/
+
+class MyHashSet {
+
+    //length of primary and nested arrs
+    private int buckets;
+    private int bucketItems;
+    private boolean [][]storage;
+    
+    //primary bucket
+    private int hash1(int key){
+        return key%buckets;
+    }
+
+    //nested array bucket
+    private int hash2(int key){
+        return key/bucketItems;
+    }
+
+    public MyHashSet() {
+        this.buckets=1000;
+        this.bucketItems=1000;
+        this.storage = new boolean[buckets][];
+    }
+
+    public void add(int key) {
+        int bucket = hash1(key);
+        //check for nested array
+        if(storage[bucket]==null){
+            //only for 0 element edge case to set 1000000 value
+            if(bucket==0){
+            storage[bucket]= new boolean[bucketItems +1];
+        }
+        else{
+            storage[bucket]=new boolean[bucketItems];
+        }
+    }
+    int bucketItem = hash2(key);
+    storage[bucket][bucketItem]=true;    
+    }
+
+    public void remove(int key) {
+        int bucket = hash1(key);
+        int bucketItem = hash2(key);
+        //check for nested array
+        if(storage[bucket]==null) return;
+        storage[bucket][bucketItem]=false;
+    }
+
+    public boolean contains(int key) {
+        int bucket =hash1(key);
+        int bucketItem = hash2(key);
+        //check for nested array
+        if(storage[bucket] == null) {
+            return false;
+        }
+        return storage[bucket][bucketItem];
+    }
+}
\ No newline at end of file
diff --git a/Problem2.java b/Problem2.java
new file mode 100644
index 00000000..2b111c5d
--- /dev/null
+++ b/Problem2.java
@@ -0,0 +1,58 @@
+// Time Complexity :O(1)
+// Space Complexity : O(n)
+// 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
+/*
+In this two arrays are used one to store values and one to track the minimum at each position.
+When I push a value, I also store the smallest value seen so far
+This will let me get the minimum element in constant time.
+time  - O(1)
+space - O(n)
+*/
+
+class MinStack {
+    int[] stack;
+    int[] minStack;
+    int top;
+
+    public MinStack() {
+        stack = new int[10000];
+        minStack = new int[10000];
+        top = -1;
+    }
+
+    public void push(int val) {
+        top++;
+        stack[top] = val;
+        // if first element, it is the minimum by default
+        if (top == 0) {
+            minStack[top] = val;
+        } //else, store the smaller value between the current element and the previous minimum so far
+        else {
+            minStack[top] = Math.min(val, minStack[top - 1]);
+        }
+    }
+
+    public void pop() {
+        top--;
+    }
+
+    public int top() {
+        return stack[top];
+    }
+
+    public int getMin() {
+        return minStack[top];
+    }
+}
+
+/**
+ * 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();
+ */
\ No newline at end of file
diff --git a/Sample.java b/Sample.java
deleted file mode 100644
index 1739a9cb..00000000
--- a/Sample.java
+++ /dev/null
@@ -1,7 +0,0 @@
-// Time Complexity :
-// Space Complexity :
-// Did this code successfully run on Leetcode :
-// Any problem you faced while coding this :
-
-
-// Your code here along with comments explaining your approach