Interview Question for Software Development Engineer Intern at Amazon.com:

This one was easy. Walk through string array with for loop, add each char to a java Set data structure. the add() method in a Java set returns false if the object is already in the set. When 'false' happens, break out of loop and return the current index. return -1 otherwise.

I'm not sure if there is a way to do this without using a temporary data structure, but this is O(n) in worst case time complexity, so it's pretty good.

public static void main(String[] args) {
        // Return the index of the first repeated character of a string.

        String s = "abdcbm";

        boolean[] map = new boolean[28];
        for (int i = 0; i < s.length(); i++) {
            if (!map[s.charAt(i) - 'a']) {
                map[s.charAt(i) - 'a'] = true;
            } else {
                System.out.println(i);
                break;
            }
        }

    }

Well, arsh, your solution will only work for lowercase letters.

How about this?

void func( char const* str)
{
    char val = 0;
    for(int i = 0; i < strlen(str); i++)
    {
          mask = 1 << str[i];
          if(mask & val)
                return i;
          else
                val |= mask;
    };

};
}

def func(str) {
    dict = {}
    l = len(str)
    for i in range(0,l):
         c = str[i]
         if c in dict:
            return dict[c]
         else:
            dict[c] = i