CareerCup

// Implement an algorithm to determine if a string has all unique characters. What if you can not use additional data structures?
// Complexity = O( n )   and space complexity O ( k ) where k is the length of the charset for ASCII  = 256
	public static boolean hasUniqueChars(char str []){
		boolean letters [] = new boolean[256];
		
		for(int i = 0; i < str.length ; i ++){
			if(str[i] == ' ') continue;
			char current = toUpperCase(str[i]);
			if(letters[current]){
				return false;
			}else{
				letters[current] = true;
			}
		}
		return true;
	}
	public static char toUpperCase(char c){
		if( c >= 'a' && c <= 'z'){
			c -= ('a' - 'A');
		}
		return c;
	}
	public static char toLowerCase(char c){
		if( c >= 'A' && c <= 'Z'){
			c += ('a'-'A');
		}
		return c;
	}

I would have same solution as you, but is array considered additional data structures?

This wont work, because it would only find 'matched' characters if they happen to be at str[i] and str[n] - if the 2 identical characters are at str[i] and str[n-1], it would not find it? Did you try st = "ganegh";?

Does this work with definition??

public static void main(String[] args) {
        
                    
    
    String a="Shirish"; int y=0;
        for(int i=0;i<a.length() && y!=0;i++)
        {
            for(int j=i+1;j<a.length();j++)
            {
                if(a.charAt(i)!=a.charAt(j))
                    y=1;
                else {y=0;}
                               

            }
        }        
         if(y==1)
                System.out.print("no unique");
            else
                System.out.print(" unique"); 
    }
    
}

Written in console application

string kelime = Console.ReadLine();
            bool varMi = true;
            for (int i = 1; i < kelime.Length; i++)
            {
                if (kelime[0]==kelime[i])
                {
                    varMi = false;
                    break;
                }
            }
            Console.WriteLine("is all characters unique: {0}", varMi);
            Console.Read();

sort the string O(nlogn).....traverse each char in string and check if next char == curr charr if yes break ....if no till u reach the end of the string return true .O(n);
total time complexity == O(nlogn)

sort the string O(nlogn).....traverse each char in string and check if next char == curr charr if yes break ....if no till u reach the end of the string return true .O(n);
total time complexity == O(nlogn)

sort the string O(nlogn).....traverse each char in string and check if next char == curr charr if yes break ....if no till u reach the end of the string return true .O(n);
total time complexity == O(nlogn)

sort the string O(nlogn).....traverse each char in string and check if next char == curr charr if yes break ....if no till u reach the end of the string return true .O(n);
total time complexity == O(nlogn)

sort the string O(nlogn).....traverse each char in string and check if next char == curr charr if yes break ....if no till u reach the end of the string return true .O(n);
total time complexity == O(nlogn)

gadha kahi ka..

/* this one is the easy approach to do this, simply scan the string from both side if found any match character then its not unique. time complexity O(n/2) i.e. O(n). */

public class uniquestring {
public static boolean test_string(String str){
int n = str.length()-1;
int i=0;
while(i<n)
{
if(str.charAt(i++)== str.charAt(n--))
return true;
}
return false;
}
public static void main(String[] args){
String st;
st = "ganesh";
System.out.println(test_string(st));
}
}