From f66a05049963d0e4c61a84a4b57ed2c9dfa75acd Mon Sep 17 00:00:00 2001
From: Vaishnavi Gawale <vaishnavigawale@mac.myfiosgateway.com>
Date: Sat, 27 Dec 2025 23:24:05 -0500
Subject: [PATCH] Done Design1

---
 MinStack.py  | 47 +++++++++++++++++++++++++++++++++++++++++++++++
 MyHashSet.py | 51 +++++++++++++++++++++++++++++++++++++++++++++++++++
 2 files changed, 98 insertions(+)
 create mode 100644 MinStack.py
 create mode 100644 MyHashSet.py

diff --git a/MinStack.py b/MinStack.py
new file mode 100644
index 00000000..4e953104
--- /dev/null
+++ b/MinStack.py
@@ -0,0 +1,47 @@
+''' Time Complexity : O(1) for all the operations
+    Space Complexity : O(n) ; where n is no of elements and we are maintaining the stack
+    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
+
+   Approach : Maintaining the one to one mapping between stack and min_stack,
+                to keep the track of previous minimums.
+'''
+
+
+class MinStack:
+
+    def __init__(self):
+        self.stack = []
+        self.min_stack = []
+        self.min=float("infinity")
+
+    def push(self, val: int) -> None:
+        self.stack.append(val)
+        if val < self.min:
+            self.min=val
+        self.min_stack.append(self.min)
+
+    def pop(self) -> None:
+        self.stack.pop()
+        self.min_stack.pop()
+        if self.min_stack:
+            self.min = self.min_stack[-1]
+        else:
+            self.min = float("infinity")
+
+        
+    def top(self) -> int:
+        return self.stack[-1]
+        
+    def getMin(self) -> int:
+        return self.min_stack[-1]
+        
+
+# Your MinStack object will be instantiated and called as such:
+# obj = MinStack()
+# obj.push(val)
+# obj.pop()
+# param_3 = obj.top()
+# param_4 = obj.getMin()
\ No newline at end of file
diff --git a/MyHashSet.py b/MyHashSet.py
new file mode 100644
index 00000000..76309976
--- /dev/null
+++ b/MyHashSet.py
@@ -0,0 +1,51 @@
+''' Time Complexity : O(1) for all the operations
+    Space Complexity : O(n) ; where n is no of elements will be added to the hashset
+    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
+
+   Approach : First, initialized the storage with primary array and boolean array, then 
+              implemented the first and second hash function. 
+'''
+
+class MyHashSet:
+
+    def __init__(self):
+        self.bucket = 1000
+        self.bucketItem = 1001
+        self.storage = [[] for i in range(self.bucket)]
+
+    def add(self, key: int) -> None:
+        
+        index = key%self.bucket
+        if self.storage[index] == []:
+            self.storage[index] = [False for i in range(self.bucketItem)]
+
+        index2 = key//self.bucketItem
+        self.storage[index][index2] = True
+        
+    def remove(self, key: int) -> None:
+        index = key%self.bucket
+        if self.storage[index]:
+            index2 = key//self.bucketItem
+            self.storage[index][index2] = False
+
+    def contains(self, key: int) -> bool:
+        index = key%self.bucket
+        index2 = key//self.bucketItem
+        if self.storage[index] != []: 
+            return self.storage[index][index2]   
+        else:
+            return False
+
+obj = MyHashSet()
+print(obj.add(1))
+print(obj.add(2))
+print(obj.contains(1))
+print(obj.contains(3))
+print(obj.add(2))
+print(obj.contains(2))
+print(obj.remove(2))
+print(obj.contains(2))