Completed Design-2

MyHashMap.py

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+# Time Complexity : 
+# - put: O(1) on average, O(n) in the worst case due to collision resolution
+# - get: O(1) on average, O(n) in the worst case if the key is in a long chain
+# - remove: O(1) on average, O(n) in the worst case for the same reason as get
+#
+# Space Complexity : O(n), where n is the number of key-value pairs stored in the hashmap.
+#
+# Did this code successfully run on Leetcode : Yes
+#
+# Any problem you faced while coding this : No significant issues; the chaining method for collision handling worked well.
+
+class MyHashMap:
+
+    def __init__(self):
+        # Initialize the hash map with a fixed size (to reduce collisions)
+        self.size = 1000  # Size of the hash table
+        self.buckets = [[] for _ in range(self.size)]  # Create empty buckets
+
+    def _hash(self, key: int) -> int:
+        # Compute the hash for a given key
+        return key % self.size
+
+    def put(self, key: int, value: int) -> None:
+        # Insert or update the key-value pair
+        index = self._hash(key)
+        # Search for the key in the corresponding bucket
+        for i, (k, v) in enumerate(self.buckets[index]):
+            if k == key:
+                # Key found; update the value
+                self.buckets[index][i] = (key, value)
+                return
+        # Key not found; append new (key, value) pair
+        self.buckets[index].append((key, value))
+
+    def get(self, key: int) -> int:
+        # Retrieve the value associated with the key
+        index = self._hash(key)
+        # Search for the key in the corresponding bucket
+        for k, v in self.buckets[index]:
+            if k == key:
+                return v  # Return the associated value
+        return -1  # Key not found
+
+    def remove(self, key: int) -> None:
+        # Remove the key-value pair if it exists
+        index = self._hash(key)
+        # Search for the key in the corresponding bucket
+        for i, (k, v) in enumerate(self.buckets[index]):
+            if k == key:
+                # Key found; remove the (key, value) pair
+                del self.buckets[index][i]
+                return
+
+# Your MyHashMap object will be instantiated and called as such:
+# obj = MyHashMap()
+# obj.put(key, value)
+# param_2 = obj.get(key)
+# obj.remove(key)

Q_Using_stacks.py

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+# Time Complexity : 
+# - push: O(1)
+# - pop: O(n) in the worst case (when the second stack is empty and we need to transfer elements)
+# - peek: O(n) in the worst case (same reason as pop)
+# - empty: O(1)
+# 
+# Space Complexity : O(n), where n is the number of elements in the queue.
+#
+# Did this code successfully run on Leetcode : Yes
+#
+# Any problem you faced while coding this : No significant issues, just ensuring proper transfer of elements between stacks.
+
+class MyQueue:
+
+    def __init__(self):
+        # Initialize two stacks: one for input and one for output
+        self.input_stack = []   # Stack to hold incoming elements
+        self.output_stack = []  # Stack to hold elements for popping
+
+    def push(self, x: int) -> None:
+        # Push the element onto the input stack
+        self.input_stack.append(x)
+
+    def pop(self) -> int:
+        # Ensure the output stack has elements to pop
+        self.peek()  # This will populate the output stack if it's empty
+        return self.output_stack.pop()
+
+    def peek(self) -> int:
+        # If the output stack is empty, transfer elements from input stack
+        if not self.output_stack:
+            while self.input_stack:
+                # Pop from input stack and push to output stack
+                self.output_stack.append(self.input_stack.pop())
+        # Return the top element of the output stack (front of the queue)
+        return self.output_stack[-1]
+
+    def empty(self) -> bool:
+        # The queue is empty if both stacks are empty
+        return not self.input_stack and not self.output_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()