jatinsharrma

'''
Recursion Function 
For 3 disk (Here S= Source, D= Destination,T= Temp)

                                                  3[S,D,T]
                                                /     |     \
                                              /       |       \
                                     2[S,T,D]      1[S,D,T]     2[T,D,S]
                                   /     |     \              /     |     \
                                 /       |       \    1[T,S,D]   1[T,D,S]   1[S,D,T]

jatinsharrma

#-------------------------------------------------------------------------------------------------------------------------
#----------------------------------------------INSERTION SORT-------------------------------------------------------------
#-------------------------------------------------------------------------------------------------------------------------


def insertion_sort(unsorted_list):
    i = len(unsorted_list)
    for num in range(i):
        for num1 in range(num):
            if unsorted_list[num] < unsorted_list[num1]:

jatinsharrma

'''-------------------------------------------------------------------------------------------------------------------------
--------------------------------------------------MERGE SORT-------------------------------------------------------------
-------------------------------------------------------------------------------------------------------------------------'''

def merge_sort(unsorted_list):
    if len(unsorted_list) == 1:
        return unsorted_list

    mid = len(unsorted_list)//2

jatinsharrma

#Find question here: https://www.hackerrank.com/challenges/designer-pdf-viewer/problem?h_r=internal-search

def designerPdfViewer(h, word):
    big = 0
    l_st = list(set(list(word)))
    for ch in l_st:
        val = ord(ch) - 97
        if h[val] > big:
            big = h[val]
    return len(word) * big

jatinsharrma

#For question refer : https://www.hackerrank.com/challenges/organizing-containers-of-balls/problem
''' logic
Suppose there are two container A and B and there are two type of balls in each container (qunatity may varry)

            Type0   Type 1
A contains [  1    ,   1  ]
B contains [  1    ,   1  ]

First i am calculating total no of balls of each type.
In above example Type0 balls are 2 and Type1 balls are 2

jatinsharrma

#For question refer: https://www.hackerrank.com/challenges/strange-code/problem

def strangeCounter(t):
    ti = 1
    val = 3
    while True:   
        new_ti = ti+val
        new_val = 2*val
        if t in range(ti, new_ti):
            temp = t - ti

jatinsharrma

# for question refer: https://www.hackerrank.com/challenges/angry-professor/problem
def angryProfessor(a, k):
    count = 0
    for stud in a:
        if stud < 1:
            count += 1
    if count >= k:
        return "NO"
    else: 
        return "YES"

jatinsharrma

#------------Recursive Solution----------------
def BinarySearch(array,target):
    return BinarySearchHelper(array,target,0, len(array)-1)

def BinarySearchHelper(array, target, left, right):

    middle = (right + left)//2 
    
    if left > right:
        return -1

jatinsharrma

print("//////////////////////////////////////////////////////////////////")
print("-------------------Fibanocci Series--------------------------------")
print("///////////////////////////////////////////////////////////////////\n\n\n")

print("-------------------Itertaive Solution------------------------------\n\n")

print("-------------------Using List--------------------------------------\n")
def fibonacciSeries1(n):
    series = [0,1]
    i = 2

jatinsharrma

# Uncomment code if you want to see what is actually going inside
'''
Documentation
(I Have numbered the steps in each method if you are not understanding any step refer to the number)
a) remove() and remove1()
    1. Stores the target node
    2. Stores the previous node to the target node
    3. Help determining where to append replacement's node after removing target node
    4. This loop spit out target node and previous node to the target node
    5. If node becomes None that means target node is not in the tree so retune false