Implement Queue using 2 stacks

design_hashmap.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
+
+class ListNode:
+    def __init__(self, key=-1, val=-1, next=None):
+        self.key = key
+        self.value = val
+        self.next = next
+
+class MyHashMap:
+
+    def __init__(self):
+        self.primaryarray = 10000
+        self.storage = [None] * self.primaryarray
+
+
+    def getKeyHash(self, key):
+        return key % self.primaryarray
+
+    def getPrevNode(self, key, node):
+        prev = None
+        curr = node
+        while curr:
+            if curr.key == key:
+                return prev
+            prev = curr
+            curr = curr.next
+        return prev
+
+    def put(self, key: int, value: int) -> None:
+        keyHash = self.getKeyHash(key)
+        if not self.storage[keyHash]:
+            self.storage[keyHash] = ListNode()
+        node = self.storage[keyHash]
+        prevNode = self.getPrevNode(key, node)
+        if not prevNode.next:
+            prevNode.next = ListNode(key, value)
+        else:
+            prevNode.next.value = value
+
+    def get(self, key: int) -> int:
+        keyHash = self.getKeyHash(key)
+        if not self.storage[keyHash]:
+            return -1
+        else:
+            node = self.storage[keyHash]
+            prev = self.getPrevNode(key, node)
+            if not prev.next:
+                return -1
+            return prev.next.value
+
+
+    def remove(self, key: int) -> None:
+        keyHash = self.getKeyHash(key)
+        if not self.storage[keyHash]:
+            return
+        node = self.storage[keyHash]
+        prevNode = self.getPrevNode(key, node)
+        if not prevNode.next:
+            return
+        prevNode.next = prevNode.next.next

queue_using_two_stacks.py

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+#Time Complexity : Amortized O(1) for push, pop, and peek operations
+# 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
+
+
+class MyQueue:
+
+    def __init__(self):
+        self.stack1 = []
+        self.stack2 = []
+
+
+    def push(self, x: int) -> None:
+        self.stack1.append(x)
+
+
+    def pop(self) -> int:
+        if self.stack2:
+            return self.stack2.pop()
+        else:
+            while self.stack1:
+                self.stack2.append(self.stack1.pop())
+            return self.stack2.pop()
+
+
+
+
+    def peek(self) -> int:
+        if self.stack2:
+            return self.stack2[-1]
+        else:
+            while self.stack1:
+                self.stack2.append(self.stack1.pop())
+            return self.stack2[-1]
+        return -1
+
+
+    def empty(self) -> bool:
+        return not self.stack1 and not self.stack2
+
+# 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()