How to check if a string contains 2 of the same character

Question

Using for loops, while loops, and/or if statements. How can you find number pairs in a string where the numbers are even indexes

I used if statements which works for a 6 character string but this would take forever if given a large string. So how can I make this code more efficient?

string = '8h8s9c'
if string[0] == string[2] or string[0] == string[4] :
   print('pair')
if string[2] == string[0] or string[2] == string[4] :
   print('pair')
else:
   print('This string contains no pair')

8h8s9c should print pair

5s7a5k should print pair

Answer 1

Since we're only worried about even indexes, we can right away cut out all the other characters.

string = string[::2]

Then check each character we care about (decimal digits) and see if it's in there twice.

if any(string.count(x)>1 for x in '0123456789'):print('pair')
else:print('This string contains no pair')

Answer 2

Super simple answer in python:

string = '8h8s9c'

string = string[::2]; #string = "889", this basically splices the string

string1 = list(string) #string1 = ['8','8','9'], separates into digits

string2 = set(string1) #string2 = {'9','8'} basically all duplicates get eliminated if any. therefore IF there are duplicates, number of elements WILL reduce.

print(len(string1) == len(string2)) # false as 3 != 2

Answer 3

 def func(s):
      l  = list(s)
      for i in s:
           if l.count(i)>1:
                return True
      return False

s = "8h8s9c"
if func(s):
     print("pair")
else:
     print("not pair")

output

pair

Answer 4

this will give you a dictionary with all pairs and # of occurrences:

from collections import Counter
counts = Counter(string)
out = {key:counts[key] for key in counts.keys() if counts[key] > 1}
if out:
    print('Pair')

output:

{'8': 2} #output of out
Pair