chore: Completed Design-1

Author: pkmaster21

design_hashset.py

@@ -0,0 +1,39 @@
+# Time Complexity : O(1)* - * could be O(n) if all n inserted values happen to fall in the same bucket, thus requiring us to scan through the entire array n-times
+# Space Complexity : O(n) - for n elements being added
+# 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 initiated an array of empty arrays so we can add new values to them by using % bucket_size. in each bucket, we add/remove the same way we would a normal array.
+
+
+class MyHashSet:
+    # Time: O(1)
+    # Space: O(1)
+    def __init__(self):
+        self.buckets = [[] for _ in range(1000)]
+
+    # Time: O(1)*
+    # Space: O(1)
+    def add(self, key: int) -> None:
+        if key not in self.buckets[key % 1000]:
+            self.buckets[key % 1000].append(key)
+
+    # Time: O(1)*
+    # Space: O(1)
+    def remove(self, key: int) -> None:
+        if key in self.buckets[key % 1000]:
+            self.buckets[key % 1000].remove(key)
+
+    # Time: O(1)*
+    # Space: O(1)
+    def contains(self, key: int) -> bool:
+        return key in self.buckets[key % 1000]
+
+
+# Your MyHashSet object will be instantiated and called as such:
+# obj = MyHashSet()
+# obj.add(key)
+# obj.remove(key)
+# param_3 = obj.contains(key)

design_minstack.py

@@ -0,0 +1,47 @@
+# Time Complexity : O(1)
+# Space Complexity : O(n)
+# Did this code successfully run on Leetcode : yes
+# Any problem you faced while coding this : no, only understanding the concept but once i understood it, coding was straightforward
+
+
+# Your code here along with comments explaining your approach
+# It look me a little bit to understand minstack but I came to the conclusion that two lists are needed for it: one to keep track of a normal stack, and one to keep track of the minimum value for constant time retrieval. Once I understood that, the rest of the functions was pretty self explanatory and straighforward
+
+
+class MinStack:
+
+    def __init__(self):
+        self.stack = []
+        self.minStack = []
+
+    # Time: O(1)
+    # Space: O(1)
+    def push(self, val: int) -> None:
+        self.stack.append(val)
+        if len(self.minStack) == 0 or val <= self.minStack[-1]:
+            self.minStack.append(val)
+
+    # Time: O(1)
+    # Space: O(1)
+    def pop(self) -> None:
+        val = self.stack.pop()
+        if val == self.minStack[-1]:
+            self.minStack.pop()
+
+    # Time: O(1)
+    # Space: O(1)
+    def top(self) -> int:
+        return self.stack[-1]
+
+    # Time: O(1)
+    # Space: 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()