# HackerRank - 30 Days Python Challenge ## 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 ```python # 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 ```python #!/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 ```python #!/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?) ```python # 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 ``` {% hint style="success" %} #### Tip: items() method for looping through dictionaries dict = {...} for key, value in dict.items(): print(key, value) {% endhint %} Recap: Looping over a list and appending filtered values (To filter out only numbers) ```python import math raw_data = [1,2,3, "nan"] filtered_data = [] for value in raw_data: #use math.isnan() if !math.isnan(value): filtered_data.append(value) ``` ## Day 9 : Recursion & Factorial of Number ```python #!/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 ```python #!/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 ```python print(len(max(bin(int(input().strip()))[2:].split('0')))) ``` Plan 3: Use Bin first to convert then count 1s ```python 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 ```python 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) ``` --- # Agent Instructions: Querying This Documentation If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question. Perform an HTTP GET request on the current page URL with the `ask` query parameter: ``` GET https://davin-hong3.gitbook.io/d/programming/hackerrank/hackerrank-30-days-python-challenge.md?ask= ``` The question should be specific, self-contained, and written in natural language. The response will contain a direct answer to the question and relevant excerpts and sources from the documentation. Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.