CRCK: Array 1.1 (easy)

Cracking the Code interview 1.1: Implement an algorithm to determine if a string has all unique characters and what if you can't use additional data structures. 

(easy level)

// First with extra data structures.
// we can use a hashtable to remember the characters and a boolean to represent existing or not.
bool isUnique(string input) {
    // if the input string size equals to 0 or 1, there is no repeats.
    if(input.size() <= 1) return true;
    bool hashMap[256] = {false};
    for(int i = 0; i < input.size(); ++i) {
        if(hashMap[input[i]]) return false;
        hashMap[input[i]] = true;
    }
    return true;
}

If we are not allowed to use extra data structure. we will then need to confirm with the interviewer the encoding ways. UniCode or UTF-8, 32, 64 ?

// we are not allowed to use extra data structure. But we can use temporary value.
// The idea is simple:
// suppose the input ranges from 'a' to 'z' 'A' to 'Z' and we have input string: "abc"
// We use an integer's bits (64 bits) to remember whether the characters has showed up before.
// "abc" will be remembered by the integer as 0000...000111;
bool isUnique(string input) {
    // same as above, we need to check the length of input.
    if(input.size() <= 1) return true;
    int characters = 0;
    for(int i = 0; i < input.size(); ++i) {
        // if the bit have been turned into '1', there is a repeat.
        if(characters & (1 << int(input[i] - 'a'))) return false;
        // otherwise, we set the bit as 1.
        characters |= (1 << int(input[i] - 'a'));
    }
    return true;
}