Code to output the first repeated character in given string?

You can do this in an easier way:

letters = 'acbdc'
found_dict = {}
for i in letters:
    if i in found_dict:
        print(i)
        break
    else:
        found_dict[i]= 1

Output: c

Here's a solution with sets, it should be slightly faster than using dicts.

letters = 'acbdc'
seen = set()

for letter in letters:
    if letter in seen:
        print(letter)
        break
    else:
        seen.add(letter)

Here is a solution that would stop iteration as soon as it finds a dup

>>> from itertools import dropwhile
>>> s=set(); next(dropwhile(lambda c: not (c in s or s.add(c)), letters))
'c'

You should use range(0, len(letters)) instead of range(0, len(letters) - 1) because range already stops counting at one less than the designated stop value. Subtracting 1 from the stop value simply makes you skip the last character of letters in this case.

Please read the documentation of range: https://docs.python.org/3/library/stdtypes.html#range

There were a few issues with your code...

1.Remove -1 from len(letters)

2.Move back one indent and do b = b + 1 even if you don't go into the if statement

3.Indent and do a = a + 1 in the first for loop.

See below of how to fix your code...

    letters = 'acbdc'
    for a in range(0, len(letters)):
        # print(letters[a])
        for b in range(0, len(letters)):
            # print(letters[b])
            if (letters[a] == letters[b]) and (a != b):
                print(b)
            b = b + 1
        a = a + 1

Nice one-liner generator:

l = 'acbdc'
next(e for e in l if l.count(e)>1)

Or following the rules in the comments to fit the "abba" case:

l = 'acbdc'
next(e for c,e in enumerate(l) if l[:c+1].count(e)>1)

If complexity is not an issue then this will work fine.

letters = 'acbdc'
found = False
for i in range(0, len(letters)-1):
    for j in range(i+1, len(letters)):
        if (letters[i] == letters[j]):
            print (letters[j])
            found = True
            break
    if (found):
        break

The below code prints the first repeated character in a string. I used the functionality of the list to solve this problem.

      def findChar(inputString): 
          list = []
          for c in inputString:
              if c in list:
                  return c
              else:
                  list.append(c) 
      return 'None'    

      print (findChar('gotgogle'))

Working fine as well. It gives the result as 'g'.

def first_repeated_char(str1):
    for index,c in enumerate(str1):
      if str1[:index+1].count(c) > 1:
        return c 
    return "None"

print(first_repeated_char("abcdabcd"))

str_24 = input("Enter the string:")
for i in range(0,len(str_24)):
    first_repeated_count = str_24.count(str_24[i])
    if(first_repeated_count > 1):
       break
print("First repeated char is:{} and character is 
      {}".format(first_repeated_count,str_24[i]))

You can use regular expressions for this task.

import re
def find_first_duplicate(input_string):
    pattern = re.compile(r'(\w)\1')
    answer = pattern.findall(input_string)
    if not answer: 
        return None
    else:
        return answer[0]

In this code:

• 1st Capturing Group \w matches any word character (equivalent to [a-zA-Z0-9_])
• \1 matches the same text as most recently matched by the 1st capturing group