Write a program that accepts a sequence of whitespace separated words as input and prints the words after removing all duplicate words and sorting them alphanumerically.
Suppose the following input is supplied to the program:
hello world and practice makes perfect and hello world again
Then, the output should be:
again and hello makes perfect practice world
In case of input data being supplied to the question, it should be assumed to be a console input.We use set container to remove duplicated data automatically and then use sorted() to sort the data.
Main author's Solution: Python 2
s = raw_input()
words = [word for word in s.split(" ")]
print " ".join(sorted(list(set(words))))My Solution: Python 3
word = input().split()
for i in word:
if word.count(i) > 1: #count function returns total repeatation of an element that is send as argument
word.remove(i) # removes exactly one element per call
word.sort()
print(" ".join(word))OR
word = input().split()
[word.remove(i) for i in word if word.count(i) > 1 ] # removal operation with comprehension method
word.sort()
print(" ".join(word))OR
word = sorted(list(set(input().split()))) # input string splits -> converting into set() to store unique
# element -> converting into list to be able to apply sort
print(" ".join(word))'''Solution by: Sukanya-Mahapatra
'''
inp_string = input("Enter string: ").split()
out_string = []
for words in inp_string:
if words not in out_string:
out_string.append(words)
print(" ".join(sorted(out_string)))Write a program which accepts a sequence of comma separated 4 digit binary numbers as its input and then check whether they are divisible by 5 or not. The numbers that are divisible by 5 are to be printed in a comma separated sequence.
Example:
0100,0011,1010,1001
Then the output should be:
1010
Notes: Assume the data is input by console.
In case of input data being supplied to the question, it should be assumed to be a console input.
Main author's Solution: Python 2
value = []
items=[x for x in raw_input().split(',')]
for p in items:
intp = int(p,2)
if not intp % 5:
value.append(p)
print ','.join(value)My Solution: Python 3
def check(x): # converts binary to integer & returns zero if divisible by 5
total,pw = 0,1
reversed(x)
for i in x:
total+=pw * (ord(i) - 48) # ord() function returns ASCII value
pw*=2
return total % 5
data = input().split(",") # inputs taken here and splited in ',' position
lst = []
for i in data:
if check(i) == 0: # if zero found it means divisible by zero and added to the list
lst.append(i)
print(",".join(lst))OR
def check(x): # check function returns true if divisible by 5
return int(x,2)%5 == 0 # int(x,b) takes x as string and b as base from which
# it will be converted to decimal
data = input().split(',')
data = list(filter(check,data)) # in filter(func,object) function, elements are picked from 'data' if found True by 'check' function
print(",".join(data))OR
data = input().split(',')
data = list(filter(lambda i:int(i,2)%5==0,data)) # lambda is an operator that helps to write function of one line
print(",".join(data))'''Solution by: nikitaMogilev
'''
data = input().split(',')
data = [num for num in data if int(num, 2) % 5 == 0]
print(','.join(data))'''Solution by: hajimalung baba
'''
print(*(binary for binary in input().split(',') if int(binary,base=2)%5==0))Write a program, which will find all such numbers between 1000 and 3000 (both included) such that each digit of the number is an even number.The numbers obtained should be printed in a comma-separated sequence on a single line.
In case of input data being supplied to the question, it should be assumed to be a console input.
Main author's Solution: Python 2
values = []
for i in range(1000, 3001):
s = str(i)
if (int(s[0])%2 == 0) and (int(s[1])%2 == 0) and (int(s[2])%2 == 0) and (int(s[3])%2 == 0):
values.append(s)
print ",".join(values)My Solution: Python 3
lst = []
for i in range(1000,3001):
flag = 1
for j in str(i): # every integer number i is converted into string
if ord(j)%2 != 0: # ord returns ASCII value and j is every digit of i
flag = 0 # flag becomes zero if any odd digit found
if flag == 1:
lst.append(str(i)) # i is stored in list as string
print(",".join(lst))OR
def check(element):
return all(ord(i)%2 == 0 for i in element) # all returns True if all digits i is even in element
lst = [str(i) for i in range(1000,3001)] # creates list of all given numbers with string data type
lst = list(filter(check,lst)) # filter removes element from list if check condition fails
print(",".join(lst))OR
lst = [str(i) for i in range(1000,3001)]
lst = list(filter(lambda i:all(ord(j)%2 == 0 for j in i), lst)) # using lambda to define function inside filter function
print(",".join(lst))'''Solution by: nikitaMogilev
'''
# map() digits of each number with lambda function and check if all() of them even
# str(num) gives us opportunity to iterate through number by map() and join()
print(','.join([str(num) for num in range(1000, 3001) if all(map(lambda num: int(num) % 2 == 0, str(num)))]))'''Solution by: hajimalung
'''
from functools import reduce
#using reduce to check if the number has only even digits or not
def is_even_and(bool_to_compare,num_as_char):
return int(num_as_char)%2==0 and bool_to_compare
print(*(i for i in range(1000,3001) if reduce(is_even_and,str(i),True)),sep=',')Write a program that accepts a sentence and calculate the number of letters and digits.
Suppose the following input is supplied to the program:
hello world! 123
Then, the output should be:
LETTERS 10
DIGITS 3
In case of input data being supplied to the question, it should be assumed to be a console input.
Main author's Solution: Python 2
s = raw_input()
d = {"DIGITS":0, "LETTERS":0}
for c in s:
if c.isdigit():
d["DIGITS"]+=1
elif c.isalpha():
d["LETTERS"]+=1
else:
pass
print "LETTERS", d["LETTERS"]
print "DIGITS", d["DIGITS"]My Solution: Python 3
word = input()
letter,digit = 0,0
for i in word:
if ('a'<=i and i<='z') or ('A'<=i and i<='Z'):
letter+=1
if '0'<=i and i<='9':
digit+=1
print("LETTERS {0}\nDIGITS {1}".format(letter,digit))OR
word = input()
letter, digit = 0,0
for i in word:
if i.isalpha(): # returns True if alphabet
letter += 1
elif i.isnumeric(): # returns True if numeric
digit += 1
print(f"LETTERS {letter}\n{digits}") # two different types of formating method is shown in both solution''' Solution by: popomaticbubble
'''
import re
input_string = input('> ')
print()
counter = {"LETTERS":len(re.findall("[a-zA-Z]", input_string)), "NUMBERS":len(re.findall("[0-9]", input_string))}
print(counter)'''Solution by: MarkisLandis
'''
sen = input("").split(" ")
alp, digit = 0, 0
for item in sen:
lst = [char for char in item]
for j in lst:
if 64 < ord(j) < 123:
alp += 1
if j.isdigit():
digit += 1
print(f"LETTERS : {alp} \n DIGITS : {digit}")'''Solution by: hajimalung
'''
#using reduce for to count
from functools import reduce
def count_letters_digits(counters,char_to_check):
counters[0] += char_to_check.isalpha()
counters[1] += char_to_check.isnumeric()
return counters
print('LETTERS {0}\nDIGITS {1}'.format(*reduce(count_letters_digits,input(),[0,0])))All the above problems are mostly string related problems. Major parts of the solution includes string releted functions and comprehension method to write down the code in more shorter form.