Print first non repetitive character of string in Python

Use such option:

def firstNonRepeat(s):
    counter = {}
    for char in s:
        if char in counter:
            counter[char] += 1
        else:
            counter[char] =  1
    for item in counter:
        if counter[item] == 1:
            return item
print(firstNonRepeat('aabccbdcbe'))

Idea is to return the first match in for loop and stop loop by returning the match. Also if the for loop will not find anything, it will return None, there is no need to write that last return.

@Zorro: Your solution is almost correct. Just do a minor change in your code and you are good to go.You have to write the return statement as it finds a character with count 1. I am writing the solution for you below:

    def firstNonRepeat(s):
        counter = {}

        for char in s:
            if char in counter:
                counter[char] += 1
            else:
                counter[char] =  1

        for char in s:
            if counter[char] == 1:
                print (char)
                return

    firstNonRepeat('aabccbdcbe')

Use break when you find the character you're looking for to stop the for loop.

for char in s:
    if counter[char] == 1:
      print char
      break

Okay so first, I may have misinterpreted what you were looking for but if you are looking for a character that does not immediately repeat ('aa') then this should work for you.

You are iterating over the entire array one and a portion. I did it like this:

def first_non_repeat(s):
        if len(s) == 1:
                return s
        for i in range(len(s)):
                if (i == 0 and s[i+1] != s[i]) or (i<len(s)-1 and s[i-1] != s[i] and s[i] != s[i+1]) or (i == len(s)-1 and s[i-1] != s[i]):
                        return s[i] # or return i
        return None

It loops over the array but only as much as it needs to in order to get the first non-repeated letter. The if statement checks if the character behind and in front of the current index don't match the one at the current index, if so it returns that letter. Note, you could also have it return the index which would potentially be more useful.

If you put in

"HHHHELLLO WORLD!"

the output from the function is 'E' or if you changed it to return index value it will return 4.

def firstNonRepeat(s):
    counter = {}
    result = []


    for char in s:
        if char in counter:
            counter[char] += 1
        else:
            counter[char] =  1
            result.append(char)

    for i in result:
        if counter[i] == 1:
            return(i)

def findFirstUniqueChar(s): d = {} for c in s: if c in d: d[c] += 1 else: d[c] = 1

for c in s:
    if d[c] == 1:
        return s.index(c)
return -1