CareerCup

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;
}

void findUniqueCharactersInString() {
		char[] s = "Write a program to find the unique char in the ascii string.".toCharArray();
		int i = 0;
		boolean[] chars = new boolean[512];
		while (i < s.length) {
			int j = i + 1;
			while (j < s.length && !chars[s[i]]) {
				if (s[i] == s[j]) {
					break;
				}
				j++;
			}
			if (j == s.length) {
				System.out.println(s[i] + " ");
			}
			chars[s[i]] = true;
			i++;
		}

}

Simple forloop in Python

#Write a program to find the unique char in the ascii string.
string = 'Federico'
uniques = []
for letter in string:
	if letter not in uniques:
		uniques.append( letter)

#output = Fedrico
print ''.join( uniques )

As the string is ASCII we can easily use a bitmap to count the occurrence of each character. But as the question asks for uniqueness, counting is not required and one bit per element should be sufficient. Anyways here is the code for general count based bitmap solution

long[] bitmap = new long[256];
	
	public void printUniqChar(char[] s) {
		for(char c:s)
			bitmap[c]++;
		for(int i=0; i<bitmap.length; i++)
			if(bitmap[i] == 1)
				System.out.printf("UNIQUE = %c\n", i);
	}

#include<iostream>
#include <map>

using namespace std;


int main(){
char arr[]="hello";

map<char,int> mymap;
map<char,int>::iterator it;
int i=0;
while(arr[i]){
if(!mymap.count(arr[i]))
mymap[arr[i]]=0;
else
mymap[arr[i]]=1;
i++;
}

for (it=mymap.begin(); it != mymap.end(); it++ )
cout << (*it).first << " => " << (*it).second << endl;


for(int i=0;i<10;i++){
if(!mymap.find(arr[i])->second)
cout<<"unique first number is:"<<arr[i]<<endl;
//break;
}

}

I've been working on the algorithm and here's what I noticed that would also work. It makes the algorithm easier to understand when you exercise it by hand:

public static boolean isUniqueChars(String str) {
        if (str.length() > 26) { // Only 26 characters
            return false;
        }
        int checker = 0;
        for (int i = 0; i < str.length(); i++) {
            int val = str.charAt(i) - 'a';
            int newValue = Math.pow(2, val) // int newValue = 1 << val
            if ((checker & newValue) > 0) return false;
            checker += newValue // checker |= newValue
        }
        return true;

When we get the value of `val` (0-25), we could either shift 1 to the right by the value of `val`, or we could use the power of 2s.

Also, for as long as the `((checker & newValue) > 0)` is false, the new checker value is generated when we sum up the old checker value and the `newValue`.