Completed PreCourse1

Exercise_1.py

@@ -1,24 +1,43 @@
+# Time Complexity : O(1) ; All stack operation perform in constant time
+# Space Complexity : O(n); n = number of elements in the stack
+# Did this code successfully run on Leetcode : - 
+# Any problem you faced while coding this : No
+
+# Your code here along with comments explaining your approach
+
 class myStack:
-  #Please read sample.java file before starting.
-  #Kindly include Time and Space complexity at top of each file
-     def __init__(self):
+  def __init__(self):
+    self.stack=[]
 
-     def isEmpty(self):
+  def isEmpty(self):  #checking if stack is Empty
+    return False if self.stack else True
 
-     def push(self, item):
+  def push(self, item):
+    self.stack.append(item)
 
-     def pop(self):
+  def pop(self):
+    if self.isEmpty():
+      print("Stack is empty; Cannot pop")
+    else:
+      return self.stack.pop() #remove and return the top element
 
+  def peek(self):
+    if self.isEmpty():
+      print("Stack is empty; Cannot peek")
+    else:
+      return self.stack[-1]
 
-     def peek(self):
-
-     def size(self):
-
-     def show(self):
+  def size(self):
+    return len(self.stack)
 
+  def show(self):
+    return self.stack
 
+
 s = myStack()
 s.push('1')
 s.push('2')
 print(s.pop())
 print(s.show())
+
+

Exercise_2.py

@@ -1,16 +1,30 @@
+# Time Complexity : O(1) ; All stack operation perform in constant time
+# Space Complexity : O(n); n = number of elements in the stack
+# Did this code successfully run on Leetcode : - 
+# Any problem you faced while coding this : No
 
+# Your code here along with comments explaining your approach
 class Node:
     def __init__(self, data):
        self.data = data
        self.next = None
 
 class Stack:
     def __init__(self):
-
+        self.head = None
+
     def push(self, data):
-
+        newnode = Node(data)
+        newnode.next = self.head
+        self.head = newnode
+
     def pop(self):
-
+        if self.head:
+            x = self.head.data
+            self.head = self.head.next
+            return x
+
+
 a_stack = Stack()
 while True:
     #Give input as string if getting an EOF error. Give input like "push 10" or "pop"

Exercise_3.py

@@ -1,8 +1,19 @@
+'''Time Complexity : O(n) ; n = number of elements in the LinkedList
+        To insert, remove or find key, we need to traverse entire linklist
+Space Complexity : O(n); 
+Did this code successfully run on Leetcode : - 
+Any problem you faced while coding this : No
+
+Your code here along with comments explaining your approach
+'''
+
 class ListNode:
     """
     A node in a singly-linked list.
     """
     def __init__(self, data=None, next=None):
+        self.data = data
+        self.next = None
 
 class SinglyLinkedList:
     def __init__(self):
@@ -17,16 +28,72 @@ def append(self, data):
         Insert a new element at the end of the list.
         Takes O(n) time.
         """
-
+        newnode = ListNode(data)
+        if not self.head:     #if Linklist is empty, make new node as first/head node
+            self.head=newnode
+        else:                #if linkList has nodes present, append newnode at end of list
+            temp = self.head
+            while temp.next:
+                temp = temp.next
+            temp.next = newnode
+
     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
+        while temp.data != key:
+            temp = temp.next
+        print("Key Found :",temp.data)
 
     def remove(self, key):
+        """
+        Remove the first occurrence of `key` in the list.
+        Takes O(n) time.
+        """       
+        if self.head.data == key:  # if key is at start node
+            x = self.head.data
+            self.head = self.head.next
+        else:
+            temp = self.head
+            while temp.data != key:
+                prev = temp
+                temp = temp.next
+            x = temp.data
+            if temp.next is None:   #if key is at last node 
+                prev.next = None
+            else:                   #if key is at middle node
+                prev.next = temp.next
+        print("Remove key:",x)    
+
+
+    def show(self):
         """
         Remove the first occurrence of `key` in the list.
         Takes O(n) time.
         """
+        l = []
+        temp = self.head
+        while temp:
+            l.append(temp.data)
+            temp = temp.next
+        print("Linked List : ", l)
+
+
+s = SinglyLinkedList()
+s.append(1)
+s.append(2)
+s.append(3)
+s.show()
+s.find(2)
+s.remove(1) #remove 1st node
+s.show()
+s.append(4)
+s.append(5)
+s.show()
+s.remove(5) #ewmove last node
+s.show()
+s.remove(3) #remove middle node
+s.show()