Done Design-2

DesignHashmap.py

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+''' Time Complexity :  Worst case for collision : O(n)
+    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
+
+   Approach : Primary staorage as Array of type Node and secondary storage as linkedlist
+'''
+class ListNode:
+    def __init__(self,key=None,value=None,next=None):
+        self.key=key
+        self.value=value
+        self.next=next
+
+class MyHashMap:
+
+    def __init__(self):
+        self.bucket=1000
+        self.storage = [ListNode() for _ in range(self.bucket)]
+
+    def put(self, key: int, value: int) -> None:
+        print("put :",key,value)
+        index = key % self.bucket
+        if self.storage[index]:
+            curr = self.storage[index]
+            while curr.next:
+                if curr.next.key == key:
+                    curr.next.value = value
+                    return
+                curr = curr.next
+            curr.next = ListNode(key, value)
+
+    def get(self, key: int) -> int:
+        print("get:",key)
+        index = key % self.bucket
+        if self.storage[index]:
+            curr = self.storage[index]
+            while curr.next:
+                if curr.next.key == key:
+                    return curr.next.value
+                curr = curr.next
+            return -1
+
+    def remove(self, key: int) -> None:
+        print("remove :",key)
+        index = key % self.bucket
+        if self.storage[index]:
+            curr = self.storage[index]
+            while curr.next:
+                if curr.next.key == key:
+                    curr.next = curr.next.next
+                    return
+                curr = curr.next
+
+
+# Your MyHashMap object will be instantiated and called as such:
+# obj = MyHashMap()
+# obj.put(key,value)
+# param_2 = obj.get(key)
+# obj.remove(key)

QueueUsingStack.py

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+''' Time Complexity : 
+        push operation - 0(1)
+        pop, peek : O(1) amortized
+    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
+
+   Approach : 1. Maintain 2 stack, inStack and minStack, push always in inStack
+              2. For pop and peek, pop elements from inStack to minStack
+'''
+
+class MyQueue:
+    def __init__(self):
+        self.in_stack=[]
+        self.out_stack=[]
+
+    def push(self, x: int) -> None:
+        self.in_stack.append(x)
+
+    def pop(self) -> int:
+        if not self.out_stack:
+            while self.in_stack:
+                self.out_stack.append(self.in_stack.pop())
+        return self.out_stack.pop()
+
+    def peek(self) -> int:
+        if not self.out_stack:
+            while self.in_stack:
+                self.out_stack.append(self.in_stack.pop())
+        return self.out_stack[-1]
+
+    def empty(self) -> bool:
+        return not self.in_stack and not self.out_stack
+
+# Your MyQueue object will be instantiated and called as such:
+# obj = MyQueue()
+# obj.push(x)
+# param_2 = obj.pop()
+# param_3 = obj.peek()
+# param_4 = obj.empty()