From 7dcb52d0fe91be49464b2daf6cd36972b2c8cb64 Mon Sep 17 00:00:00 2001
From: akarshchandra <95599941+akarshchandra@users.noreply.github.com>
Date: Thu, 4 Jun 2026 19:14:41 -0500
Subject: [PATCH 1/2] Create Problem1.py

---
 Problem1.py | 54 +++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 54 insertions(+)
 create mode 100644 Problem1.py

diff --git a/Problem1.py b/Problem1.py
new file mode 100644
index 00000000..fc175f49
--- /dev/null
+++ b/Problem1.py
@@ -0,0 +1,54 @@
+# Time Complexity : all O(1) although pop/peek is O(n) worst case average O(1) since only when outstack is completeley empty we need to transfer content
+# Space Complexity :O(n)
+# Did this code successfully run on Leetcode : yes
+# Any problem you faced while coding this : no
+
+# we can keep on pushhing to instack and only transfer first time to out/peek i.e when it becomes empty other timer can be poping from out stack dince value there any how means / needs to pop first irrespective of what is being added to in stack  
+
+
+class MyQueue:
+
+    def __init__(self):
+        self.instack =  []
+        self.outstack = []
+    
+    def transferInStackContents(self):
+        items = len(self.instack)
+        while items > 0:
+            self.outstack.append(self.instack.pop())
+            items = items - 1
+        return
+
+    def push(self, x: int) -> None:
+        self.instack.append(x)
+
+    def pop(self) -> int:
+        if not self.outstack:
+            self.transferInStackContents()
+        
+        if self.outstack:
+            return self.outstack.pop()
+        
+        return 0
+
+    def peek(self) -> int:
+        if not self.outstack:
+            self.transferInStackContents()
+        
+        if self.outstack:
+            return self.outstack[-1]
+
+        return 0
+        
+
+    def empty(self) -> bool:
+            return not self.instack and not self.outstack
+        
+
+
+# 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()

From 2c3df548f6f265fe0753afe723ec80b60e8578cc Mon Sep 17 00:00:00 2001
From: akarshchandra <95599941+akarshchandra@users.noreply.github.com>
Date: Thu, 4 Jun 2026 19:15:11 -0500
Subject: [PATCH 2/2] Create Problem2.py

---
 Problem2.py | 83 +++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 83 insertions(+)
 create mode 100644 Problem2.py

diff --git a/Problem2.py b/Problem2.py
new file mode 100644
index 00000000..f3fe09c6
--- /dev/null
+++ b/Problem2.py
@@ -0,0 +1,83 @@
+# Time Complexity : all 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
+# Although we can do this with double hasung since i already implemented it i will use #linear chaining approach here 
+
+# if i equal divide space for  priomary array and secondary array as in double hashing  i will have to iterate over 1000 items to remove , get calls to reduce this i will increase the primary array size to 10K with a hash func to find bucket like key % 10K and then use single link lnked list as secondary data structure at max one bucket or arr location can have 100 items at max worst case scenario 100 iters to find the element which on average is still O(1) comparef to n of 10M
+
+
+
+class MyHashMap:
+
+    class Node:
+        def __init__(self,key,val,ptr=None):
+            self.key=key
+            self.val=val
+            self.next=ptr
+
+    def __init__(self):
+        self.primSize = 10000
+        self.arr = [None] * self.primSize
+    
+    def find(self,key,ptr) -> Node:
+        prev=ptr
+        while  prev.next and prev.next.key != key:
+            prev = prev.next
+        return prev
+
+    def hash(self,key):
+        return key % 10000
+
+    def put(self, key: int, value: int) -> None:
+
+        idx = self.hash(key)
+
+        if not self.arr[idx]:
+            dummy = self.Node(-1,-1)
+            self.arr[idx] = dummy
+        
+        """important this i may make mitake i added dummy in if but both the path after that and if already a chain exists both end up here so i should call find here just after writinf if in the same mood of new chain with just dummy i can't just add to dummy"""
+
+        prev = self.find(key,self.arr[idx])
+
+        if prev.next:
+            """if prev.next is not null means we found the key in the # chain since while loop runs till prev.next has the key or prev.next is null"""
+            prev.next.val = value
+            return
+        
+        # here prev.next is None means key not found add as a new node in the end
+        prev.next = self.Node(key,value)
+        return
+        
+    def get(self, key: int) -> int:
+        idx = self.hash(key)
+        if not self.arr[idx]:
+            return -1
+        prev = self.find(key,self.arr[idx])
+        if not prev.next:
+            return -1
+        return prev.next.val
+        
+
+    def remove(self, key: int) -> None:
+        idx = self.hash(key)
+        if not self.arr[idx]:
+            return
+        prev = self.find(key,self.arr[idx])
+        if not prev.next:
+            return
+        curr = prev.next
+        prev.next = prev.next.next
+        curr.next = None
+        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)