Add HashSet and MinStack solutions Aditya

design-1.py

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+# Problem 1: Design HashSet
+# Time Complexity: O(1) for all operations - add, remove, contains
+# Space Complexity: O(n) n is number of unique keys
+# Did this code successfully run on Leetcode: YES
+# Any problem you faced while coding this: Faced error in some syntax
+
+class MyHashSet(object):
+    def __init__(self):
+        self.initial_bucket = 1000
+        self.secondary_bucket = 1000
+        self.storage = [None] * self.initial_bucket
+
+    def hash_func1(self, key):
+        return key % self.initial_bucket
+
+    def hash_func2(self, key):
+        return key // self.secondary_bucket
+
+    def add(self, key):
+        bucket = self.hash_func1(key)
+        internal_bucket = self.hash_func2(key)
+        if self.storage[bucket] is None:
+            if bucket == 0:
+                self.storage[bucket] = [False] * (self.secondary_bucket + 1)
+            else:
+                self.storage[bucket] = [False] * self.secondary_bucket
+        self.storage[bucket][internal_bucket] = True
+
+    def remove(self, key):
+        bucket = self.hash_func1(key)
+        internal_bucket = self.hash_func2(key)
+        if self.storage[bucket] is None:
+            return
+        self.storage[bucket][internal_bucket] = False
+
+    def contains(self, key):
+        bucket = self.hash_func1(key)
+        internal_bucket = self.hash_func2(key)
+        if self.storage[bucket] is None:
+            return False
+        return self.storage[bucket][internal_bucket]
+
+
+# Problem 2: Min Stack
+# Time Complexity: O(1) for all operations - push, pop, top, getMin
+# Space Complexity: O(n) 
+# Did this code successfully run on Leetcode: YES, but needed to take reference from solution to understand the approach
+# Any problem you faced while coding this: Understanding the two stack approach
+
+class MinStack:
+    def __init__(self):
+        self.main_stack = []
+        self.min_stack = []
+
+    def push(self, val: int) -> None:
+        self.main_stack.append(val)
+        if not self.min_stack or val <= self.min_stack[-1]:
+            self.min_stack.append(val)
+        else:
+            self.min_stack.append(self.min_stack[-1])
+
+    def pop(self) -> None:
+        self.main_stack.pop()
+        self.min_stack.pop()
+
+    def top(self) -> int:
+        return self.main_stack[-1]
+
+    def getMin(self) -> int:
+        return self.min_stack[-1]

min-stack.py

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+# Problem 2: Min Stack
+# Time Complexity: O(1) for all operations - push, pop, top, getMin
+# Space Complexity: O(n) 
+# Did this code successfully run on Leetcode: YES, but needed to take reference from solution to understand the approach
+# Any problem you faced while coding this: Understanding the two stack approach
+
+class MinStack:
+    def __init__(self):
+        self.main_stack = []
+        self.min_stack = []
+
+    def push(self, val: int) -> None:
+        self.main_stack.append(val)
+        if not self.min_stack or val <= self.min_stack[-1]:
+            self.min_stack.append(val)
+        else:
+            self.min_stack.append(self.min_stack[-1])
+
+    def pop(self) -> None:
+        self.main_stack.pop()
+        self.min_stack.pop()
+
+    def top(self) -> int:
+        return self.main_stack[-1]
+
+    def getMin(self) -> int:
+        return self.min_stack[-1]