design 2 completed

Sample.py

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+#Time Complexity : Amortized O(1) per operation (push, pop, peek, empty).
+#					Worst case for pop/peek: O(n) when elements are transferred from inst → outst.
+# 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
+
+
+# Your code here along with comments explaining your approach
+
+class MyQueue:
+
+    def __init__(self):
+        self.inst = []
+        self.outst = []
+
+    def push(self, x: int) -> None: #O(1)
+        self.inst.append(x)
+
+    def pop(self) -> int: #O(1)
+        if self.empty():
+        	return -1
+        self.peek() # ensure outst has the front element
+        return self.outst.pop()
+
+
+    def peek(self) -> int: #O(1)
+        if not self.outst:  # if outst is empty, transfer from inst
+        	while self.inst:
+        		self.outst.append(self.inst.pop())
+        return self.outst[-1]
+
+    def empty(self) -> bool: #O(1)
+    	return not self.inst and not self.outst
+
+
+
+