# Array Pair SumArray Pair Sum

Given an integer array, output all the unique pairs that sum up to a specific value k. So the input:

``sum_arr_uniq_pairs([1,2,2,3,4,1,1,3,2,1,3,1,2,2,4,0],5) ``

would return 2 pairs:
(2,3)  (1,4)

# Find a missing element in an array/list

Consider an array of non-negative integers. A second array is formed by shuffling the elements of the first array and deleting a random element. Given these two arrays, find which element is missing in the second array. Here is an example input, the first array is shuffled and the number 5 is removed to construct the second array.
Input:

`` find_missing_ele([1,2,3,4,5,6,7],[3,7,2,1,4,6]) ``

Output:5 is the missing numberStack class implementation

# Implement basic stack operations (LIFO)

push() – Push an element in a stack pop()- POP an element from top of the stackpeek() – Just peek into top element of the stack (don’t perform any operation)Queue class implementation

# Implement basic Queue operations (FIFO)

enqueue – adding a element to the queue dequeue – removing an element from the queueDeque (DECK) class implementation

# Implement basic operation in deque (Add and remove elements both at front and rear)

``addFront()``

Add an element at the front

``addRear()``

Add an element at the rear

``removeFront()``

Remove from front

``removeRear()``

Remove from rear

# Balance parentheses using stack/list

Given a string of opening and closing parentheses, check whether it’s balanced. We have 3 types of parentheses: round brackets: () square brackets: [] curly brackets: {}. Assume that the string doesn’t contain any other character than these, no spaces words or numbers. As a reminder, balanced parentheses require every opening parenthesis to be closed in the reverse order opened. For example ‘([])’ is balanced but ‘([)]’ is not. Algo will take a string as the input string and will return boolean (TRUE/FALSE) Examples:

``````print (check_parentheses_match('([])'))
print (check_parentheses_match('[](){([[[]]])'))
``````

# Queue with 2 stack implementation

This is a classic problem. We need to use the basic characteristics of the stack (popping out elements in reverse order) will make a queue. Example:

``````# Create a object of the class
qObj = QueueWith2Stack()
# Add an element
qObj.enqueue(1)
# Add another element
qObj.enqueue(2)
# Add more element
qObj.enqueue(4)
# Add more element
qObj.enqueue(8)
# Remove item
print (qObj.dequeue())
# Remove item
print (qObj.dequeue())
# Remove item
print (qObj.dequeue())
# Remove item
print (qObj.dequeue())
``````

# Singly Linked List class implementation

Implement basic skeleton for a Singly Linked List Example:

``````# Added node
# Set the pointers
a.nextnode = bb.nextnode = c
print (a.value)
print (b.value)
print (c.value)
# Print using class
print (a.nextnode.value)``````

# Doubly Linked List class implementation

Implement basic skeleton for a Doubly Linked List Example:

``````
# Set the pointers
# setting b after a (a before b)
b.prev_node = a
a.next_node = b
# Setting c after a
b.next_node = c
c.prev_node = b
print (a.value)
print (b.value)
print (c.value)
# Print using class
print (a.next_node.value)
print (b.next_node.value)
print (b.prev_node.value)
print (c.prev_node.value)
``````

# Reverse a linked list implementation

The aim is to write a function to reverse a Linked List in place. The function will take in the head of the list as input and return the new head of the list. Example:

``````
# Create a Linked List
a.nextnode = b
b.nextnode = c
c.nextnode = d
print (a.nextnode.value)
print (b.nextnode.value)
print (c.nextnode.value)
# Call the reverse()
print (d.nextnode.value)
print (c.nextnode.value)
print (b.nextnode.value)
``````

# Linked list Nth to the last node

The aim is a function that takes a head node and an integer value n and then returns the nth to last node in the linked list. Example:

``````
# Create a Linked List

a.nextnode = b
b.nextnode = c
c.nextnode = d
d.nextnode = e

print (a.nextnode.value)
print (b.nextnode.value)
print (c.nextnode.value)
print (d.nextnode.value)

# This would return the node d with a value of 4, because its the 2nd to last node.
target_node = LinkedListNode().nth_to_last_node(2, a)
print (target_node.value)
# Ans: d=4

``````

Check GitHub for the full working code.

I will keep adding more problems/solutions.

Stay tuned!

Ref:  The inspiration of implementing DS in Python is from this course

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