diff --git a/Problem1.py b/Problem1.py
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+# #  Time Complexity : O(1) for insert, and O(n) for remove and contains
+# #  Space Complexity : O(n)
+# #  Did this code successfully run on Leetcode : YES
+# #  Any problem you faced while coding this :
+
+
+# #  Your code here along with comments explaining your approach
+# class MyHashSet:
+
+#     # HashSet is an array of arrays lenght 10000. Constraint 0 <= key <= 10^6
+#     def __init__(self):
+#         self.n = 10000
+#         self.arr = [[] for _ in range(self.n)]
+
+#     # Hash function is mod of the length of the array 10000
+#     # to find the array index in which a key will go, we will mod the key by 10000
+#     # then check if the key already exist in the array at the given index and insert
+#     # Complexity: O(1)
+#     def add(self, key: int) -> None:
+#         index = key % self.n
+#         if key not in self.arr[index]:
+#             self.arr[index].append(key)
+
+#     # find the array index in which the key could be
+#     # if present in the array, remove it
+#     # Complexity: O(n)
+#     def remove(self, key: int) -> None:
+#         index = key % self.n
+#         if key in self.arr[index]:
+#             self.arr[index].remove(key)
+
+#     # Complexity: O(n)
+#     def contains(self, key: int) -> bool:
+#         index = key % self.n
+#         return key in self.arr[index]
+
+
+# # Your MyHashSet object will be instantiated and called as such:
+# # obj = MyHashSet()
+# # obj.add(key)
+# # obj.remove(key)
+# # param_3 = obj.contains(key)
+
+
+
+# Option 2 which uses 2 hash functions to achieve O(1) time complexity
+class MyHashSet:
+
+    def __init__(self):
+        self.n = 1000
+        self.arr = [[] for _ in range(self.n)]
+
+    # Uses 2 hash function to reduce time complexity
+    # Complexity: O(1)
+    def add(self, key: int) -> None:
+        bucket = key % self.n
+        bucketIndex = key // self.n # 2nd hash function gives a unique index in the 2nd array
+        if len(self.arr[bucket]) == 0: #init the 2nd array with default values as False
+            # 1000000 will fall at index 1000, and our arrays only go to 999
+            # hence the 1st bucket(0) will have 1 extra value 
+            self.arr[bucket] = [False for _ in range(self.n + 1 if bucket == 0 else self.n)]
+        
+        # set the key's position to True when we insert into the array
+        self.arr[bucket][bucketIndex] = True
+
+    # Complexity: O(1)
+    def remove(self, key: int) -> None:
+        # to remove we simply compute the index for both arrays for the key
+        # then set the 2nd array value to False to mark it as removed
+        bucket = key % self.n
+        bucketIndex = key // self.n
+        if len(self.arr[bucket]) > 0:
+            self.arr[bucket][bucketIndex] = False
+        
+    # Complexity: O(1)
+    def contains(self, key: int) -> bool:
+        bucket = key % self.n
+        bucketIndex = key // self.n
+        if len(self.arr[bucket]) > 0:
+            return self.arr[bucket][bucketIndex] # simply return the value in the 2nd array
+        else:
+            return False
+
+
+
+# Your MyHashSet object will be instantiated and called as such:
+obj = MyHashSet()
+obj.add(1)
+obj.add(2)
+print(obj.contains(1))
+print(obj.contains(3))
+obj.add(2)
+print(obj.contains(2))
+obj.remove(2)
+print(obj.contains(2))
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diff --git a/Problem2.py b/Problem2.py
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+class MinStack:
+    
+    # we maintain 2 arrays, one for the stack, and one for all our minimums with each push
+    def __init__(self):
+        self.stack = []
+        self.minStack = []
+
+    # when pushing on top of stack, also push the min of minStack and new val to minStack
+    # this allows O(1) retain our minimum with every push
+    # Stack will grow as follows:
+    # Stack [ 5, 6, 8, 4, 7, 3 ]
+    # MinStack [ 5, 5, 5, 4, 4, 3]
+    # Complexity: O(1)
+    def push(self, val: int) -> None:
+        self.stack.append(val)
+        minimum = min(val, self.minStack[-1] if self.minStack else val)
+        self.minStack.append(minimum)
+
+    # pop from both stacks so that we always have previous min in the minStack
+    # Complexity: O(1)
+    def pop(self) -> None:
+        self.minStack.pop()
+        self.stack.pop()
+
+    # return top of Stack
+    # Complexity: O(1)
+    def top(self) -> int:
+        if len(self.stack) > 0:
+            return self.stack[-1]
+
+    # Complexity: O(1)
+    def getMin(self) -> int:
+        return self.minStack[-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()
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