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

Question: 1st non-repeating character in a string

Method 1: making a count array

a = "GGGiniiiiGinaaaPrrrottiijayi"
count = [0]*256
for ch in a: count[ord(ch)] +=1
for ch in a :
    if( count[ord(ch)] == 1 ):
        print(ch)
        break

Method2: List comprehension

# 1st  non repeating character
pal = [x for x in a if a.count(x) == 1][0]
print(pal)

Method 3: dictionary

d={}
for ch in a: d[ch] = d.get(ch,0)+1
aa = sorted(d.items(), key = lambda ch  :ch[1])[0][0]
print(aa)

Answer 2

In Java, 1)Create the Character count hashmap. For each character,if there is no value stored in the character,set it to 1. Else increment the value of the character by 1 .

2)Then I Scanned the String for each character. Return character if the count in hashmap is 1 . If no character has count 1 ,then return null.

package com.abc.ridhi;
import java.util.HashMap;
import java.util.Scanner;

public class FirstNonRepeated {

    public static void main(String[] args) {
        System.out.println(" Please enter the input string :");
        Scanner in = new Scanner(System.in);
        String s = in.nextLine();
        char c = firstNonRepeatedCharacter(s);
        System.out.println("The first non repeated character is :  " + c);
    }

    public static Character firstNonRepeatedCharacter(String str) {
    HashMap<Character, Integer> map1 = new HashMap<Character, Integer>();
        int i, length;
        Character c;
        length = str.length(); 
        // Scan string and build hashmap
        for (i = 0; i < length; i++) {
            c = str.charAt(i);
            if (map1.containsKey(c)) {
                // increment count corresponding to c
                map1.put(c, map1.get(c) + 1);
            } else {
                map1.put(c, 1);
            }
        }
        // Search map in order of string str

        for (i = 0; i < length; i++) {
            c = str.charAt(i);
            if (map1.get(c) == 1){
                return c;
        }
        }
        return null;
    }
}

Answer 3

The following code traverses through the string only once. It's a simple and easy solution for the problem but at the same time the space complexity is compromised.

def firstNonRepeating(a):
    inarr = []
    outarr = []

    for i in a:
        #print i
        if not i in inarr:
            inarr.append(i)
            outarr.append(i)
        elif i in outarr:
            outarr.remove(i)
        else:
            continue
    return outarr[0]

a = firstNonRepeating(“Your String”)
print a

Answer 4

My solution. I can't speak to how efficient it is; I think it runs in n^2 time.

>>> def fst_nr(s):
...     collection = []
...     for i in range(len(s)):
...             if not s[i] in collection and not s[i] in s[i+1:]:
...                     return s[i]
...             else:
...                     collection+=[s[i]]
... 
>>> fst_nr("teeter")
'r'
>>> fst_nr("floccinaucinihilipilification")
'u'
>>> fst_nr("floccinacinihilipilification")
'h'
>>> fst_nr("floccinaciniilipilification")
'p'
>>> fst_nr("floccinaciniiliilification")
't'
>>> fst_nr("floccinaciniiliilificaion")
>>> 

Any advice for a humble Stack noob?

Answer 5

Since we need the first non-repeating character, it is better to use OrderedDict from collections because dict does not guarantee the order of the keys

from collections import OrderedDict
dict_values = OrderedDict()
string = "abcdcd"

for char in string:
   if char in dict_values:
   dict_values[char] += 1
else:
    dict_values[char] = 1

for key,value in dict_values.items():
    if value == 1:
       print ("First non repeated character is %s"%key)
       break
    else:
       pass

Answer 6

def firstnonrecur(word):

    for c in word:
        if word.count(c) == 1:
            print(c)
            break


firstnonrecur('google')

How about this?