diff --git a/Sample.java b/Sample.java
index 1739a9cb..bf6ae167 100644
--- a/Sample.java
+++ b/Sample.java
@@ -1,7 +1,57 @@
-// Time Complexity :
-// Space Complexity :
-// Did this code successfully run on Leetcode :
+// Time Complexity : O(1)
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : Yes
 // Any problem you faced while coding this :
-
+// In push fucntion calculating minVal, I did mistakely poped instared of peek
+// It emptied the Stack when single element is present
 
 // Your code here along with comments explaining your approach
+class MinStack {
+    //we will store two vals in int[]
+    //First one current value, second one min element till now
+    Stack<int[]> stack;
+    public MinStack() {
+        stack = new Stack<int[]>();
+    }
+
+    public void push(int val) {
+        if(stack.isEmpty())
+        {
+            stack.push(new int[]{val,val});
+            return;
+        }
+        //Calc min value till now
+        int minVal = stack.peek()[1];
+        //Calc curr min by comparinging minVal and incoming val
+        int currMin = Math.min(minVal,val);
+        //push the {val,currMin} into stack
+        stack.push(new int[]{val,currMin});
+    }
+
+    public void pop() {
+        //O(1) time
+        stack.pop();
+
+    }
+
+    public int top() {
+        //O(1) time
+        return stack.peek()[0];
+
+    }
+
+    public int getMin() {
+        //O(1) time
+        return stack.peek()[1];
+
+    }
+}
+
+/**
+ * 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();
+ */