HackerRank - 30 Days Python Challenge

My goal is to commit to 2 challenges a day for this 30 day HackerRank python challenge, while simultaneously studying for OSCP and doing boxes (Target : 1hr per day)

Day 6 : Lets Review

So this challenge mainly reviews knowledge of strings, loops and accessing indexes

Task

Given a string, , of length that is indexed from to , print its even-indexed and odd-indexed characters as space-separated strings on a single line (see the Sample below for more detail).

Note: is considered to be an even index.

Example

s = adbecf

Print abc def

Solution

https://thispointer.com/python-reverse-a-list-sub-list-or-list-of-list-in-place-or-copy/

# Enter your code here. Read input from STDIN. Print output to STDOUT
# First input is number of test cases, T
T = int(input())

#Now iterate exactly T times for per input string S

for i in range (0,T):
    #take twice input string
    S=input()
    # Now this is the function for slicing indexes of even + odd
    print(S[0::2] + " " + S[1::2])
    # first [0::2] takes index zero of string S then jumps by twos to the end and so on..
    
    

Day 7 : Arrays

Task

Given an array, , of integers, print 's elements in reverse order as a single line of space-separated numbers.

Example

A = [1,2,3,4]

Print 4 3 2 1. Each integer is separated by one space.

Solution 1 - use slicing with negative step then join

#!/bin/python3

import math
import os
import random
import re
import sys



if __name__ == '__main__':
    n = int(input().strip())

    arr = list(map(int, input().rstrip().split()))
    #to reverse theres a few ways, one is to call from -1 steps (the last element to first)
    x = arr[::-1]
    # this gives us an array of reverse n [2, 3, 4, 1]
    y=' '.join(str(e) for e in x)
    #must use str to convert int (you cannot join int)
    print(y)
    # the above returns the array reversed, now to convert to string with spaces
    
    
    

Solution 2 - for loop

#!/bin/python3

import math
import os
import random
import re
import sys



if __name__ == '__main__':
    n = int(input().strip())

    arr = list(map(int, input().rstrip().split()))
    empty=[]
    # loop from 0 to len -1 size of list
    for i in range(0,len(arr)):
        #append item at -1 to list
        #basically u wanna count -1, -2, -3 ...
        #so if i start from 0, 1 > apply algo such that 0, 1 turns to -1, -2
        empty.append(arr[-(i+1)])
    print(' '.join(str(e) for e in empty))
    
    

Day 8 : Dictionaries & Key-value map

Task

Given n names and phone numbers, assemble a phone book that maps friends' names to their respective phone numbers. You will then be given an unknown number of names to query your phone book for. For each name queried, print the associated entry from your phone book on a new line in the form name=phoneNumber; if an entry for is not found, print Not found instead.

The criteria was for the phone book to be of: Dictionary, Map or HashMap data structure

Input format:

3 #cater dictionary length to n provided
sam 99912222
tom 11122222
harry 12299933
sam
edward
harry #these queries can go unlimited

Solution

Used first to map to dictionary structure, next perform while True if loop for input in dict (is the name query a key in the dictionary?)

# Enter your code here. Read input from STDIN. Print output to STDOUT
n = int(input())
# only takes first line, number of dict items = 3

# PART 1: CREATE DICT 
dict={}
for x in range(0,int(n)):
    s = input()
    # s now reads the dict items names and num
    y=s.split(" ")
    #multiple duo arrays with [name1, num1] and so on
    dict[y[0]]=y[1]
#print(dict) #now we have a dict of the names and nums linked ready
 
#PART 2: QUERIES HANDLING
while True:
    try:
        inp=input()
        if inp in dict: #if names queries are keys in dict
            print(inp + "=" + dict[inp])
        else:
            print("Not found")
    except EOFError:
        break

Tip: items() method for looping through dictionaries

dict = {...}

for key, value in dict.items():

print(key, value)

Recap: Looping over a list and appending filtered values (To filter out only numbers)

import math

raw_data = [1,2,3, "nan"]
filtered_data = []
for value in raw_data:
#use math.isnan(<insert to be filtered>)
    if !math.isnan(value):
        filtered_data.append(value)
        
        

Day 9 : Recursion & Factorial of Number

#!/bin/python3
import math
import os
import random
import re
import sys
#main
def factorial(n):
    # intialise
    result = 1
    for i in range(2,n+1):
        result = result*i
        #input 3 = 3*2*1 (we exclude 2 to cut process time)
    return result

#the below is provided to process input
if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    n = int(input().strip())

    result = factorial(n)

    fptr.write(str(result) + '\n')

    fptr.close()

Day 10 : Converting Base10 to Base2

Given a base10 number, convert to base2 and find the maximum number of consecutive ones '1'.

Plan 1:

• counter of current no. of consecutive 1 (if current counter > maximum counter, set it to maximum counter value)

• counter of maximum consecutive 1

• once hit zero, reset counter to zero

#!/bin/python3
import sys

n = int(input().strip())
# first intialize counters
current_consecutive_ones=0
max_consecutive_ones=0
while n>0:
    remainder = n%2
    if remainder == 1:
        current_consecutive_ones += 1 #now to let max_consecutive catchup to current counter      
        if current_consecutive_ones > max_consecutive_ones:
            current_consecutive_ones == max_consecutive_ones
    else:  #remainder ==0 (cant be anything except 0 or 1 when divided by 2
        current_consecutive_ones = 0 #reset to zero
    n = n//2 
    #// is floor division, 9//2 = 4

print(max_consecutive_ones)

Plan 2 : Convert > bin > max > len

print(len(max(bin(int(input().strip()))[2:].split('0'))))

Plan 3: Use Bin first to convert then count 1s

num = int(input())
convert = bin(num) # binary numbers 1 or 0
current = 0
maximum = 0

for i in convert:
            if i =='1':
                        current +=1
            else:
                        current = 0 # reset to zero
                        maximum_count = max(maximum, current)
print(max(maximum,current))

Plan 4: Use append the remainders to a list, count number of 1s

if __name__ == '__main__':
    n = int(input())
    rmd = []
    while n > 0:
        rm = n % 2
        n = n//2
        rmd.append(rm)
    
    count,result = 0,0 # intialise
    
    for i in range(0,len(rmd)): #iterate thru list for ones
        if rmd[i] == 0:
            count = 0
        else:
            count +=1
            result = max(result,count)
    print(result)