Find the first non-repeated character in a string

Question

I read of a job interview question to write some code for the following:

Write an efficient function to find the first nonrepeated character in a st

Answer 1

from collections import defaultdict
word="googlethis"
dici=defaultdict(int)

#build up dici with counts of characters
for a in word:
    if dici[a]:
        dici[a]+=1
for a in word:
    if didic[a] < 2:
        return a

wouldn't that work?

Answer 2

input_str = "interesting"
#input_str = "aabbcc"
#input_str = "aaaapaabbcccq"


def firstNonRepeating(param):
    counts = {}        
    for i in range(0, len(param)):

    # Store count and index repectively
        if param[i] in counts:
            counts[param[i]][0] += 1
        else:
            counts[param[i]] = [1, i]

    result_index = len(param) - 1
    for x in counts:
        if counts[x][0] == 1 and result_index > counts[x][1]:
            result_index = counts[x][1]
    return result_index

result_index = firstNonRepeating(input_str)
if result_index == len(input_str)-1:
    print("no such character found")
else:
    print("first non repeating charater found: " + input_str[result_index])

Answer 3

In [1033]: def firstNonRep(word):
   ......:     c = collections.Counter(word)
   ......:     for char in word:
   ......:         if c[char] == 1:
   ......:             return char
   ......:         

In [1034]: word="googlethis"

In [1035]: firstNonRep(word)
Out[1035]: 'l'

EDIT: If you want to implement the same thing without using helpers like Counter:

def firstNonRep(word):
    count = {}
    for c in word:
        if c not in count:
            count[c] = 0
        count[c] += 1
    for c in word:
        if count[c] == 1:
            return c

Answer 4

sorted(word,key=lambda x:(word.count(x),word.index(x)) )[0]

I think or DSM's also consice

next(c for c in word if word.count(c) == 1) 

which is marginally more efficient

>>> word = "teeter"
>>> sorted(word,key=lambda x:(word.count(x),word.index(x)) )[0]
'r'
>>> word = "teetertotter"
>>> sorted(word,key=lambda x:(word.count(x),word.index(x)) )[0]
'o'
>>> word = "teetertotterx"
>>> sorted(word,key=lambda x:(word.count(x),word.index(x)) )[0]
'o'

Answer 5

Three lines of python code to get the solution:

word="googlethis"
processedList = [x for x in word if word.count(x)==1]
print("First non-repeated character is: " +processedList[0])

Or,

word="googlethis"
print([x for x in word if word.count(x)==1][0])

Answer 6

# I came up with another solution but not efficient comment?
str1 = "awbkkzafrocfbvwcqbb"
list1 = []
count = 0
newlen = len(str1)
find = False


def myfun(count, str1, list1, newlen):

    for i in range(count, newlen):

        if i == 0:

            list1.append(str1[i])

        else:
            if str1[i] in list1:
                str1 = str1.translate({ord(str1[i]): None})
                print(str1)
                newlen = len(str1)
                count =0
                i = count
                list1.pop()
                myfun(count,str1,list1,newlen)
            else:
                pass

    if str1.find(list1[0], 1, len(str1)) != -1 :
        pass
    else:
        print(list1[0]+" is your first non repeating character")
        exit()


myfun(count, str1, list1, newlen)