String unique characters [duplicate] - string

Question

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Determining a string has all unique characters without using additional data structures and without the lowercase characters assumption
(7 answers)
Closed 8 years ago.
Implement an algorithm to determine if a string has all unique characters. What if you can not use additional data structures?

Answer 1

If you can use a little auxiliary memory, then a small array of bits (indexed by the numerical code of the character) is all you need (if your characters are 4-byte Unicode ones you'll probably want a hashmap instead;-). Start with all bits at 0: scan the string from the start -- each time, you've found a duplicate if the bit corresponding to the current character is already 1 -- otherwise, no duplicates yet, set that bit to 1. This is O(N).
If you can't allocate any extra memory, but can alter the string, sorting the string then doing a pass to check for adjacent duplicates is the best you can do, O(N log N).
If you can't allocate extra memory and cannot alter the string, you need an O(N squared) check where each character is checked vs all the following ones.

Answer 2

we can do it by assigning a prime number to every character.. and multiply it for every character found. then on every character check that if the value is divisible by the number assigned to that character or not..

Answer 3

answer in c program
int is_uniq(char *str)
{
int i = 0, flag = 0, value = 0;
for(i = 0; i < strlen(str); i++) {
value = str[i] - 'a';
if(flag & (1 << value)) {
return 0;
}
flag |= (1 << value);
}
return 1;
}

Answer 4

for each character in the string
if any subsequent character matches it
fail
succeed

Answer 5

One possible solution - You could extract the string into an array of characters, sort the array, and then loop over it, checking to see if the next character is equal to the current one.

Answer 6

the prime method described by partik (based on this theorem). It's O(N).
# one prime per letter
primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101]
starting_byte = ?a.ord
primes_product = 1
ARGV[0].each_byte do |byte|
current_prime = primes[byte - starting_byte]
if primes_product % current_prime == 0
puts "Not unique"
exit
else
primes_product = primes_product * current_prime
end
end
puts "Unique"

Answer 7

I came on this thread for the similar question and ended up with the following solution in C#:
var data = "ABCDEFGADFGHETFASAJUTE";
var hash = new Dictionary<char, int>();
foreach (char c in data)
{
if (hash.ContainsKey(c))
{
hash[c] += 1;
}
else
{
hash.Add(c, 1);
}
}
var Characters = hash.Keys.ToArray();
var Frequencies = hash.Values.ToArray();

Answer 8

import java.io.*;
public class uniqueChar
{
boolean checkUniqueChar(String strin)
{
int m;
char []str=strin.toCharArray();
java.util.Arrays.sort(str);
for(int i=0;i<str.length-1;i++)
{
if(str[i]==str[i+1])
return false;
}
return true;
}
public static void main(String argv[]) throws IOException
{
String str;
System.out.println("enter the string\n");
InputStreamReader in=new InputStreamReader(System.in);
BufferedReader bin=new BufferedReader(in);
str=bin.readLine();
System.out.println(new uniqueChar().checkUniqueChar(str));
}
}

Answer 9

public static boolean isUniqueChars(String str) {
int checker = 0;
for (int i = 0; i < str.length(); ++i) {
int val = str.charAt(i) - 'a';
if ((checker & (1 << val)) > 0)
return false;
checker |= (1 << val);
}
return true;
}