A simple way is to go through all values in% with%, check that the current value belongs to the a list, which will be the output, and as long as it belongs, draw a new number between 1 and 11 until it does not is present in the list and then add it. In Python, it would look like this:
from random import randint
a = [1, 2, 3, 4, 1, 3]
b = []
for value in a:
while value in b:
value = randint(1, 11)
b.append(value)
print b
The result would be:
[1, 2, 3, 4, 8, 9]
That is, indexes 4 and 5 were identified as duplicates and other values were drawn in this case 8 and 9 respectively.
See working at Ideone .
In fact, this logic will work well with more than one repeated value. See:
from random import randint
a = [1, 2, 3, 4, 1, 1, 1, 1, 1]
b = []
for value in a:
while value in b:
value = randint(1, 11)
b.append(value)
print b
Generating result:
[1, 2, 3, 4, 7, 11, 8, 6, 10]
See working at Ideone .
However, if the list of values has more than 11 elements, it will take an hour for the program to try to draw a value and never find a valid one, since all values between 1 and 11 belong to the list, getting trapped infinitely in the loop b - starting from the premise that the list of values will initially contain only values between 1 and 11. Thus a check on the length of the list should be made to avoid such problem:
from random import randint
a = [1, 2, 3, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
b = []
if len(a) <= 11:
for value in a:
while value in b:
value = randint(1, 11)
b.append(value)
print b
else:
print "A lista 'a' deve conter no máximo 11 elementos."
See working at Ideone .
