Input Format
A single integer denoting .
Constraints
Output Format
Print lines where each line (in the range ) contains the respective decimal, octal, capitalized hexadecimal, and binary values of . Each printed value must be formatted to the width of the binary value of .
Sample Input
17
Sample Output
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
10 12 A 1010
11 13 B 1011
12 14 C 1100
13 15 D 1101
14 16 E 1110
15 17 F 1111
16 20 10 10000
17 21 11 10001 String Formatting - Hacker Rank Solution
We can solve this challenge using the .format tool.
"{0:{width}d} {0:{width}o} {0:{width}X} {0:{width}b}".format(i,width = width)
- is the index indicating which format argument should be placed in that spot.
- width indicates the padding space between two rows.
- , , and converts the string into decimal, octal, hexadecimal, and binary format, respectively.
Note: Capital is used to print the hexadecim
def print_formatted(number):
ReplyDelete# your code goes here
for i in range(1,number+1):
l=len('{0:b}'.format(number))
l=int(l)
print('{0:{width}d} {0:{width}o} {0:{width}X} {0:{width}b}'.format(i,width=l))
if __name__ == '__main__':
n = int(input())
print_formatted(n)
No width variable in your code
Deleteignore
Deletedef print_formatted(number):
ReplyDeletewidth = len('{:b}'.format(number))
for i in range(1,number+1):
print(str.rjust(str(i),width),str.rjust(str(oct(i)[2:]),width),str.rjust(str(hex(i).upper()[2:]),width),str.rjust(str(bin(i)[2:]),width))
if __name__ == '__main__':
n = int(input())
print_formatted(n)
len('{0:b}'.format(number))
ReplyDeleteplease explain this
0 indicates the "number"
Deleteb converts the "number" into binary
def print_formatted(number):
ReplyDelete# your code goes here
for i in range(1,number+1):
print("{} {} {} {}".format(i,oct(i)[1:],hex(i)[2:],bin(i)[2:]))
def print_formatted(number):
ReplyDelete# your code goes here
for i in range(1,number+1):
print("{} {} {} {}".format(i,oct(i)[1:],hex(i)[2:],bin(i)[2:]))
def print_formatted(number):
ReplyDelete# your code goes here
for i in range(1,number+1):
print("{} {} {} {}".format(i,oct(i)[1:],hex(i)[2:],bin(i)[2:]))
Hi ,
ReplyDeleteThanks for the code.
Eventhough , the output is same as the Expected one.
The cases are not passed.
Could you please tell me , What exactly means hex(i)[2:]?
seems issue related to the some padding spaces.
Thanks in advance.
hex is typecasting the int(i) into it's hexadecimal value and then [2:] is slicing the string after the index 2
Deletedef print_formatted(number):
ReplyDelete# your code goes here
for i in range(1, n+1):
l=len('{0:b}'.format(number))
l=int(l)
print('{0:{1}d} {0:{1}o} {0:{1}X} {0:{1}b}'.format(i,l))
if __name__ == '__main__':
n = int(input())
print_formatted(n)
def print_formatted(number):
ReplyDelete# your code goes here
global l1, l2, l3, l4
spac = " "
for i in range(1, number + 1):
oct1 = str(oct(i))[2:]
hex1 = str(hex(i))[2:].upper()
bin1 = str(bin(i))[2:]
# print(f"{i} {oct1} {hex1} {bin1}")
l1 = (len(str(i)))
l2= (len(oct1))
l3 = (len(hex1))
l4 = (len(bin1))
# print(f"{l1,l2,l3,l4}")
for i in range(1, number + 1):
oct1 = str(oct(i))[2:]
hex1 = str(hex(i))[2:].upper()
bin1 = str(bin(i))[2:]
# print(f"{i} {oct1} {hex1} {bin1}")
l11 = (len(str(i)))
l22 = (len(oct1))
l33 = (len(hex1))
l44 = (len(bin1))
# print(f"{l11, l22, l33, l44}")
s1 = l4 - l11
s2 = l4 - l22
s3 = l4 - l33
s4 = l4 - l44
print(f"{s1 *spac}{i} {s2 *spac}{oct1} {s3 *spac}{hex1} {s4 *spac}{bin1}")
if __name__ == '__main__':
n = int(input())
print_formatted(n)