solution to both problems

01-Design-Hashset.py

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+# Problem 1: Design HashSet (https://leetcode.com/problems/design-hashset/)
+
+# Time Complexity : O(1) for add, remove, and contains operations
+# Space Complexity : O(n) where n is the number of unique keys 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
+# We will use a 2D list (array of arrays) to implement the HashSet.
+# The first hash function will determine the bucket (outer array index) and the second hash function
+# will determine the position within that bucket (inner array index). 
+
+class MyHashSet:
+
+    def __init__(self):
+        self.buckets = 1000
+        self.bucketItems = 1000
+        self.storage = [None] * self.buckets
+
+    def hash1(self, key: int) -> int:
+        return key % self.buckets
+
+    def hash2(self, key: int) -> int:
+        return key // self.bucketItems
+
+    def add(self, key: int) -> None:
+        bucket = self.hash1(key)
+        bucketItem = self.hash2(key)
+
+        if self.storage[bucket] == None:
+            # Handle the special case for bucket 0 (to allow key=10^6)
+            if bucket == 0:
+                self.storage[bucket] = [False] * (self.bucketItems + 1)
+            else:
+                self.storage[bucket] = [False] * self.bucketItems
+
+        self.storage[bucket][bucketItem] = True
+
+    def remove(self, key: int) -> None:
+        bucket = self.hash1(key)
+        bucketItem = self.hash2(key)
+
+        if self.storage[bucket] == None:
+            return
+        self.storage[bucket][bucketItem] = False
+
+    def contains(self, key: int) -> bool:
+        bucket = self.hash1(key)
+        bucketItem = self.hash2(key)
+
+        if self.storage[bucket] == None:
+            return False
+        return self.storage[bucket][bucketItem]
+
+
+# Your MyHashSet object will be instantiated and called as such:
+# obj = MyHashSet()
+# obj.add(key)
+# obj.remove(key)
+# param_3 = obj.contains(key)

02-Design-MinStack.py

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+# Problem 2: Design MinStack (https://leetcode.com/problems/min-stack/) 
+# Time Complexity : O(1) for push, pop, top, and getMin operations
+# Space Complexity : O(n) where n is the number of elements in 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
+# We will use two stacks: one for the main stack operations and another to keep track of the minimum elements.
+# The min stack will only push a new element when it is less than or equal to the current minimum.
+
+class MinStack:
+
+    def __init__(self):
+        self.st = []
+        self.minSt = []
+        self.Min = int(sys.maxsize)
+        self.minSt.append(self.Min)
+
+    def push(self, val: int) -> None:
+        if val <= self.Min:
+            self.Min = val
+        self.st.append(val)
+        self.minSt.append(self.Min)
+
+    def pop(self) -> None:
+        self.st.pop()
+        self.minSt.pop()
+        self.Min = self.minSt[-1]
+
+    def top(self) -> int:
+        return self.st[-1]
+
+    def getMin(self) -> int:
+        return self.Min
+
+
+# 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()