Design-1 Completed!

Sample.py

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+# Problem 1 : Implementing Hashset Using Double Hashing
+
+# First and Second has functions were chosen in such a way that they ensure the collisions are uniform.
+
+# Time Complexity : 
+#   add()     -> O(1)
+#   remove()  -> O(1)
+#   contains()-> O(1)
+# Space Complexity : O(N)
+# Did this code successfully run on Leetcode : Yes
+class MyHashSet:
+
+    def __init__(self):
+        self.buckets=1000
+        self.bucketItems=1000
+        self.storage= [None]*self.buckets
+
+    def hash1(self, key: int) -> int:
+        return int(key%1000)
+
+    def hash2(self, key: int) -> int:
+        return int(key/1000)
+
+    def add(self, key: int) -> None:
+        idx1=self.hash1(key)
+        if self.storage[idx1] is None:
+            if idx1==0:
+                self.storage[idx1]=[False]*(self.bucketItems+1)
+            else:
+                self.storage[idx1]=[False]*self.bucketItems
+        idx2=self.hash2(key)
+        self.storage[idx1][idx2]= True
+
+    def remove(self, key: int) -> None:
+        idx1=self.hash1(key)
+        if self.storage[idx1] is None:
+            return
+        idx2=self.hash2(key)
+        self.storage[idx1][idx2]=False
+
+    def contains(self, key: int) -> bool:
+        idx1=self.hash1(key)
+        if self.storage[idx1] is None:
+            return False
+        idx2=self.hash2(key)
+        if self.storage[idx1][idx2] == True:
+            return True
+        else:
+            return False
+
+
+# Problem 2 : Min-Stack
+
+# A separate array was created just for storing the minimum elements, so O(1) time for getting minimum element was possible
+
+# Time Complexity : 
+#   push()    -> O(1)
+#   pop()     -> O(1)
+#   top()     -> O(1)
+#   getMin()  -> O(1)
+# Space Complexity : O(N)
+# Did this code successfully run on Leetcode : Yes
+class MinStack:
+
+    def __init__(self):
+        self.stack=[]
+        self.minstack=[]
+
+    def push(self, val: int) -> None:
+        if not self.minstack or val <= self.minstack[-1]:
+            self.minstack.append(val)
+        self.stack.append(val)
+
+    def pop(self) -> None:
+        if self.stack:
+            v=self.stack.pop()
+            if v == self.minstack[-1]:
+                self.minstack.pop()
+
+
+    def top(self) -> int:
+        if self.stack:
+            return self.stack[-1]
+
+    def getMin(self) -> int:
+        return self.minstack[-1]
+
+