Pre-course 1

Exercise_1.py

@@ -1,20 +1,28 @@
+# Time complexity: O(1)
+# Space complexity: O(n)
 class myStack:
   #Please read sample.java file before starting.
   #Kindly include Time and Space complexity at top of each file
      def __init__(self):
+        self.q = deque()  #initializing an empty deque
 
      def isEmpty(self):
+       return len(self.q) == 0  #returns true when stack is empty
 
      def push(self, item):
+       self.q.append(item) #adds a value to the end which is the top of the stack
 
      def pop(self):
-
+       return self.q.pop()  #removes last value that is inserted      
 
      def peek(self):
+       return self.q[-1] #returns the last(top most) value   
 
      def size(self):
+       return len(self.q) # returns no.of elements in the stack
 
      def show(self):
+       return list(self.q) #return all elements in the stack
 
 
 s = myStack()

Exercise_2.py

@@ -1,32 +1,62 @@
+# Time Complexity : O(1)
+# Space Complexity : O(n)
+# O(n)
+
+# Did this code successfully run on Leetcode : No, I did not solve this problem in leetcode
+
+# Any problem you faced while coding this : Took sometime to write the code by myself but understood by taking help
 
 class Node:
     def __init__(self, data):
-       self.data = data
-       self.next = None
+        self.data = data
+        self.next = None
 
+
 class Stack:
     def __init__(self):
+        self.head = None  # top of stack
 
     def push(self, data):
+        # create new node
+        new_node = Node(data)
+
+        # point new node to current head
+        new_node.next = self.head
+
+        # update head to new node
+        self.head = new_node
 
     def pop(self):
+        # check if stack is empty
+        if self.head is None:
+            return None
+
+        # store current head data
+        popped = self.head.data
+
+        # move head to next node
+        self.head = self.head.next
+
+        return popped
 
 a_stack = Stack()
 while True:
-    #Give input as string if getting an EOF error. Give input like "push 10" or "pop"
     print('push <value>')
     print('pop')
     print('quit')
+
     do = input('What would you like to do? ').split()
-    #Give input as string if getting an EOF error. Give input like "push 10" or "pop"
     operation = do[0].strip().lower()
+
     if operation == 'push':
         a_stack.push(int(do[1]))
+
     elif operation == 'pop':
         popped = a_stack.pop()
         if popped is None:
             print('Stack is empty.')
         else:
             print('Popped value: ', int(popped))
+
     elif operation == 'quit':
         break

Exercise_3.py

@@ -1,32 +1,79 @@
+# Time Complexity : append -> O(n), find -> O(n),remove -> O(n)
+
+# Space Complexity : O(n)
+
+# Did this code successfully run on Leetcode : Did not solve it on leet code, but I tested it on online compiler
+
+# Any problem you faced while coding this : Felt it bit tricky but finally understood and I'm poor at handling edge cases
+
+
 class ListNode:
     """
     A node in a singly-linked list.
     """
     def __init__(self, data=None, next=None):
-
+        self.data = data        # store value in node
+        self.next = next        # pointer to next node
+
+
 class SinglyLinkedList:
     def __init__(self):
         """
         Create a new singly-linked list.
         Takes O(1) time.
         """
-        self.head = None
+        self.head = None        # initialize head as None
 
     def append(self, data):
         """
         Insert a new element at the end of the list.
         Takes O(n) time.
         """
+        new_node = ListNode(data)   # create new node
 
+        if self.head is None:       # if list is empty
+            self.head = new_node    # make new node as head
+            return
+
+        temp = self.head            # start from head
+
+        while temp.next:            # traverse till last node
+            temp = temp.next
+
+        temp.next = new_node        # attach new node at end
+
     def find(self, key):
         """
         Search for the first element with `data` matching
         `key`. Return the element or `None` if not found.
         Takes O(n) time.
         """
+        temp = self.head            # start from head
 
+        while temp:                 # traverse list
+            if temp.data == key:    # check if value matches
+                return temp         # return node if found
+            temp = temp.next
+
+        return None                 # return None if not found
+
     def remove(self, key):
         """
         Remove the first occurrence of `key` in the list.
         Takes O(n) time.
         """
+        temp = self.head            # start from head
+        prev = None                 # keep track of previous node
+
+        while temp:
+            if temp.data == key:    # if value matches
+
+                if prev is None:   # if node to delete is head
+                    self.head = temp.next  # move head forward
+                else:
+                    prev.next = temp.next  # bypass current node
+
+                return             # exit after deletion
+
+            prev = temp            # move prev forward
+            temp = temp.next       # move temp forward