How to find first non-repetitive character from a string? - php

Question

I've spent half day trying to figure out this and finally I got working solution.
However, I feel like this can be done in simpler way.
I think this code is not really readable.
Problem: Find first non-repetitive character from a string.
$string = "abbcabz"
In this case, the function should output "c".
The reason I use concatenation instead of $input[index_to_remove] = ''
in order to remove character from a given string
is because if I do that, it actually just leave empty cell so that my
return value $input[0] does not not return the character I want to return.
For instance,
$str = "abc";
$str[0] = '';
echo $str;
This will output "bc"
But actually if I test,
var_dump($str);
it will give me:
string(3) "bc"
Here is my intention:
Given: input
while first char exists in substring of input {
get index_to_remove
input = chars left of index_to_remove . chars right of index_to_remove
if dupe of first char is not found from substring
remove first char from input
}
return first char of input
Code:
function find_first_non_repetitive2($input) {
while(strpos(substr($input, 1), $input[0]) !== false) {
$index_to_remove = strpos(substr($input,1), $input[0]) + 1;
$input = substr($input, 0, $index_to_remove) . substr($input, $index_to_remove + 1);
if(strpos(substr($input, 1), $input[0]) == false) {
$input = substr($input, 1);
}
}
return $input[0];
}

Answer 1

<?php
// In an array mapped character to frequency,
// find the first character with frequency 1.
echo array_search(1, array_count_values(str_split('abbcabz')));

Answer 2

Python:
def first_non_repeating(s):
for i, c in enumerate(s):
if s.find(c, i+1) < 0:
return c
return None
Same in PHP:
function find_first_non_repetitive($s)
{
for($i = 0; i < strlen($s); $i++) {
if (strpos($s, $s[i], $i+1) === FALSE)
return $s[i];
}
}

Answer 3

Pseudocode:
Array N;
For each letter in string
if letter not exists in array N
Add letter to array and set its count to 1
else
go to its position in array and increment its count
End for
for each position in array N
if value at potition == 1
return the letter at position and exit for loop
else
//do nothing (for clarity)
end for
Basically, you find all distinct letters in the string, and for each letter, you associate it with a count of how many of that letter exist in the string. then you return the first one that has a count of 1
The complexity of this method is O(n^2) in the worst case if using arrays. You can use an associative array to increase it's performance.

Answer 4

1- use a sorting algotithm like mergesort (or quicksort has better performance with small inputs)
2- then control repetetive characters
non repetetive characters will be single
repetetvives will fallow each other
Performance : sort + compare
Performance : O(n log n) + O(n) = O(n log n)
For example
$string = "abbcabz"
$string = mergesort ($string)
// $string = "aabbbcz"
Then take first char form string then compare with next one if match repetetive
move to the next different character and compare
first non-matching character is non-repetetive

Answer 5

This can be done in much more readable code using some standard PHP functions:
// Count number of occurrences for every character
$counts = count_chars($string);
// Keep only unique ones (yes, we use this ugly pre-PHP-5.3 syntax here, but I can live with that)
$counts = array_filter($counts, create_function('$n', 'return $n == 1;'));
// Convert to a list, then to a string containing every unique character
$chars = array_map('chr', array_keys($counts));
$chars = implode($chars);
// Get a string starting from the any of the characters found
// This "strpbrk" is probably the most cryptic part of this code
$substring = strlen($chars) ? strpbrk($string, $chars) : '';
// Get the first character from the new string
$char = strlen($substring) ? $substring[0] : '';
// PROFIT!
echo $char;

Answer 6

$str="abbcade";
$checked= array(); // we will store all checked characters in this array, so we do not have to check them again
for($i=0; $i<strlen($str); $i++)
{
$c=0;
if(in_array($str[$i],$checked)) continue;
$checked[]=$str[$i];
for($j=$i+1;$j<=strlen($str);$j++)
{
if($str[$i]==$str[$j])
{
$c=1;
break;
}
}
if($c!=1)
{
echo "First non repetive char is:".$str[$i];
break;
}
}

Answer 7

This should replace your code...
$array = str_split($string);
$array = array_count_values($array);
$array = array_filter($array, create_function('$key,$val', 'return($val == 1);'));
$first_non_repeated_letter = key(array_shift($array));
Edit: spoke too soon. Took out 'array_unique', thought it actually dropped duplicate values. But character order should be preserved to be able to find the first character.

Answer 8

Here's a function in Scala that would do it:
def firstUnique(chars:List[Char]):Option[Char] = chars match {
case Nil => None
case head::tail => {
val filtered = tail filter (_!=head)
if (tail.length == filtered.length) Some(head) else firstUnique(filtered)
}
}
scala> firstUnique("abbcabz".toList)
res5: Option[Char] = Some(c)
And here's the equivalent in Haskell:
firstUnique :: [Char] -> Maybe Char
firstUnique [] = Nothing
firstUnique (head:tail) = let filtered = (filter (/= head) tail) in
if (tail == filtered) then (Just head) else (firstUnique filtered)
*Main> firstUnique "abbcabz"
Just 'c'
You can solve this more generally by abstracting over lists of things that can be compared for equality:
firstUnique :: Eq a => [a] -> Maybe a
Strings are just one such list.

Answer 9

Can be also done using array_key_exists during building an associative array from the string. Each character will be a key and will count the number as value.
$sample = "abbcabz";
$check = [];
for($i=0; $i<strlen($sample); $i++)
{
if(!array_key_exists($sample[$i], $check))
{
$check[$sample[$i]] = 1;
}
else
{
$check[$sample[$i]] += 1;
}
}
echo array_search(1, $check);