Completed Design-2 problems

MyHashMap.py

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+"""
+Approach:
+
+Implement Hashmap using list of list of size k (any constant value) - call it buckets (outer list)
+
+Store key,value pair as a list in a bucket at an index
+Index is determined using hash function
+Store bucket in buckets list
+
+buckets (example):
+
+[
+[],                                       bucket at index 0
+[[1,24], [1001,5], [2001, 7],......],     bucket at index 1
+[],                                       bucket at index 2
+[[3,8], [1003,27], [8003, 4],.....],      bucket at index 3
+.
+.
+.
+[],                                       bucket at index 999
+]
+
+Time Complexity: O(1)
+Space = O(k) + O(n)
+      ~ O(n) where n is the number of pairs stored
+
+"""
+
+class MyHashMap:
+
+    def __init__(self):
+        self.size = 1000
+        self.buckets = [[] for _ in range(self.size)]
+
+    def _hash(self, key: int) -> int:
+        return key%self.size    
+
+    def put(self, key: int, value: int) -> None:
+        # find bucket where we can store given key,value pair using hash function
+        bucket = self.buckets[self._hash(key)]
+
+        for pair in bucket:
+            # if key already exists in bucket (e.g. [1,5]), and we have push(1,9) update key with new value instead of adding duplicate key
+            if pair[0] == key:
+                pair[1] = value
+                return
+
+        # if new key: add pair in bucket
+        bucket.append([key,value])
+
+
+    def get(self, key: int) -> int:
+        # find bucket where we can find the value
+        bucket = self.buckets[self._hash(key)]
+
+        # once you find bucket, iterate through pairs and find value
+        for k,v in bucket:
+            if k == key:
+                return v
+        # element not found
+        return -1
+
+
+    def remove(self, key: int) -> None:
+        # find bucket from which we need to remove pair
+        bucket = self.buckets[self._hash(key)]
+
+        for i, (k,v) in enumerate(bucket):
+            # if you found key, delete key value pair from bucket
+            if k == key:
+                bucket.pop(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)

MyQueue.py

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+"""
+Approach:
+1. Use 2 stacks - stack1 for push operations, stack2 for pop/peek operations.
+   (Stack is LIFO, but reversing a stack's order using a second stack gives FIFO behavior - that's how queue is simulated.)
+2. push(): just append to stack1 - O(1), no reordering needed yet.
+3. pop()/peek(): if stack2 is empty, transfer all elements from stack1 into stack2 
+   (this reverses their order, so the oldest element ends up on top of stack2).
+   Then pop()/peek() from stack2 directly.
+
+Dry run:
+push(1), push(2), push(3), push(4):
+stack1 = [1,2,3,4]   (4 is top/most recent)
+stack2 = []
+
+pop():
+stack2 is empty -> transfer all from stack1 to stack2:
+stack1 = []
+stack2 = [4,3,2,1]   (1 is now on top - it was the FIRST pushed, now first to come out)
+return stack2.pop() -> returns 1 (correct FIFO order - oldest element out first)
+
+Next pop():
+stack2 = [4,3,2] (already has elements, no transfer needed)
+return stack2.pop() -> returns 2
+
+empty():
+return True only if BOTH stack1 and stack2 are empty
+
+Time Complexity: 
+  - push(): O(1) always
+  - pop()/peek(): O(1) amortized
+  - empty(): O(1)
+
+Space Complexity: O(n) - both stacks combined hold at most n elements total (n = elements pushed)
+"""
+class MyQueue:
+
+    def __init__(self):
+        self.stack1 = []
+        self.stack2 = []
+
+
+    def push(self, x: int) -> None:
+        self.stack1.append(x)
+
+
+    def pop(self) -> int:
+        if not self.stack2:
+            while self.stack1:
+                self.stack2.append(self.stack1.pop()) 
+        return self.stack2.pop()
+
+
+    def peek(self) -> int:
+        if not self.stack2:
+            while self.stack1:
+                self.stack2.append(self.stack1.pop()) 
+        return self.stack2[-1]
+
+
+    def empty(self) -> bool:
+        if len(self.stack1) == 0 and len(self.stack2) == 0:
+            return True
+        return False
+
+
+
+# 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()