Any mistake in below code?

Question

Some test cases passed...

a=input()
b=0
for i in range(0,len(a)):
for j in range(1,len(a)):
if a[i]==a[j]:
b=b+1
if b>0:
print('false')
elif b==0:
print('true')

Answer 1

SoloProg ,

sorting is not required when using a set. we can just compare the length of the string with the length of the set:

a = input().lower()
print(len(a) == len(set(a)))

Answer 2

Arun Venkatesh.N ,

the issue of your code is that the second for loop with the range() always starts generating the same start value. see the modification in the code:

a=input()
b=0
for i in range(0, len(a)):
for j in range(i + 1, len(a)): # <<<
if a[i] == a[j]:
b=b+1
if b > 0:
print('false')
elif b == 0:
print('true')

a further impovement could be to use break when variable `b` is incremented. be aware that we have to exit both loops.
this procedre reduces the number of iterations.

Answer 3

i = 1, j = 1
a[1] == a[1] true , lead to wrong.

Instead of 1 in range(), try i+1
or add condition i!=j must

Answer 4

Thank you it's working

Answer 5

I solved it but with a diifferent way

a = input().lower()
print ( str( sorted(a) == sorted(set(a)) ).lower() )

Answer 6

Lothar ,

This is a solution using LEN to make the code work.

s = input()
b = len(s) == len(set(s))
print ( str(b).lower() )