Completed Design1

Hashset.py

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+#// Time Complexity : Averge of O(1) unless everything ends up in one bucket.
+#// Space Complexity : O(N)
+#// Did this code successfully run on Leetcode : Yes
+#// Any problem you faced while coding this : Add difficulty in converting add function concept to code.
+
+#// Your code here along with comments explaining your approach
+#Treating every number as a grid coordinate -> 2D (double hashing)
+#Create the secondary bucket values when needed instead of ccreating at start
+#Using boolean switch insted of actual values(True/false).
+
+class MyHashSet(object):
+
+    def __init__(self):
+        self.bucket = 1000
+        self.items_per_bucket = 1000
+        self.storage = [None] * self.bucket
+
+    def primary_hash(self,key):
+        return key % self.bucket
+
+    def secondary_hash(self,key):
+        return key //self.items_per_bucket
+
+
+    def add(self, key):
+        """
+        :type key: int
+        :rtype: None
+        """
+        primary_index = self.primary_hash(key)
+        secondary_index = self.secondary_hash(key)
+
+        if self.storage[primary_index] is None:
+            if primary_index == 0:
+                size = self.items_per_bucket + 1
+            else:
+                size = self.items_per_bucket
+            self.storage[primary_index] = [False] * size
+        self.storage[primary_index][secondary_index] = True
+
+    def remove(self, key):
+        """
+        :type key: int
+        :rtype: None
+        """
+        primary_index = self.primary_hash(key)
+        secondary_index = self.secondary_hash(key)
+
+        #if bucket exist then only the key
+        if self.storage[primary_index] is not None:
+            self.storage[primary_index][secondary_index] = False
+
+    def contains(self, key):
+        """
+        :type key: int
+        :rtype: bool
+        """
+        primary_index = self.primary_hash(key)
+        secondary_index = self.secondary_hash(key)
+
+        return self.storage[primary_index] is not None and self.storage[primary_index][secondary_index]
+
+
+
+# Your MyHashSet object will be instantiated and called as such:
+# obj = MyHashSet()
+# obj.add(key)
+# obj.remove(key)
+# param_3 = obj.contains(key)

Minstack.py

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+#// Time Complexity : O(1)
+#// Space Complexity : O(n)
+#// 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
+#2 stacks - 1 for storing actual pushed values,
+# 2 - for storing minimum values at each level,
+# these 2 stacks must have 1:1 mapping
+class MinStack(object):
+
+    def __init__(self):
+        self.stack = []
+        self.min_stack = []
+
+    def push(self, val):
+        """
+        :type val: int
+        :rtype: None
+        """
+        #updating min value
+        if not self.min_stack:
+            self.min_stack.append(val)
+        else:
+            top = self.min_stack[-1]
+            if val < top:
+                self.min_stack.append(val)
+            else:
+                self.min_stack.append(top)
+        self.stack.append(val)
+
+
+    def pop(self):
+        """
+        :rtype: None
+        """
+
+        self.stack.pop()
+        self.min_stack.pop()
+
+
+    def top(self):
+        """
+        :rtype: int
+        """
+        return self.stack[-1]
+
+
+    def getMin(self):
+        """
+        :rtype: 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()