Determining a string has all unique characters without using additional data structures and without the lowercase characters assumption

问题

This is one of the questions in the Cracking the Coding Interview book by Gayle Laakmann McDowell:

Implement an algorithm to determine if a string has all unique characters. What if you can not use additional data structures?

The author wrote:

We can reduce our space usage a little bit by using a bit vector. We will assume, in the below code, that the string is only lower case 'a' through 'z'. This will allow us to use just a single int.

The author has this implementation:

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;
}

Let's say we get rid of the assumption that "the string is only lower case 'a' through 'z'". Instead, the string can contain any kind of character—like ASCII characters or Unicode characters.

Is there a solution as efficient as the author's (or a solution that comes close to being as efficient as the author's)?

回答1:

for the asccii character set you can represent the 256bits in 4 longs: you basically hand code an array.

public static boolean isUniqueChars(String str) {
    long checker1 = 0;
    long checker2 = 0;
    long checker3 = 0;
    long checker4 = 0;
    for (int i = 0; i < str.length(); ++i) {
        int val = str.charAt(i);
        int toCheck = val / 64;
        val %= 64;
        switch (toCheck) {
            case 0:
                if ((checker1 & (1L << val)) > 0) {
                    return false;
                }
                checker1 |= (1L << val);
                break;
            case 1:
                if ((checker2 & (1L << val)) > 0) {
                    return false;
                }
                checker2 |= (1L << val);
                break;
            case 2:
                if ((checker3 & (1L << val)) > 0) {
                    return false;
                }
                checker3 |= (1L << val);
                break;
            case 3:
                if ((checker4 & (1L << val)) > 0) {
                    return false;
                }
                checker4 |= (1L << val);
                break;
        }            
    }
    return true;
}

You can use the following code to generate the body of a similar method for unicode characters:

static void generate() {
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < 1024; i++) {
        sb.append(String.format("long checker%d = 0;%n", i));
    }
    sb.append("for (int i = 0; i < str.length(); ++i) {\n"
            + "int val = str.charAt(i);\n"
            + "int toCheck = val / 64;\n"
            + "val %= 64;\n"
            + "switch (toCheck) {\n");
    for (int i = 0; i < 1024; i++) {
        sb.append(String.format("case %d:\n"
                + "if ((checker%d & (1L << val)) > 0) {\n"
                + "return false;\n"
                + "}\n"
                + "checker%d |= (1L << val);\n"
                + "break;\n", i, i, i));
    }
    sb.append("}\n"
            + "}\n"
            + "return true;");
    System.out.println(sb);
}

回答2:

You only need one line... well less than one line actually:

if (str.matches("((.)(?!.*\\1))*"))

this uses a negative look ahead to assert that each character is not repeated later in the string.

This approach a time complexity of O(n^2), because for all n characters in the input, all characters that follow (there are n of those) are compared for equality.

回答3:

I think we need a general and practical definition of "additional data structures". Intuitively, we don't want to call every scalar integer or pointer a "data structure", because that makes nonsense of any prohibition of "additional data structures".

I propose we borrow a concept from big-O notation: an "additional data structure" is one that grows with the size of the data set.

In the present case, the code quoted by the OP appears to have a space requirement of O(1) because the bit vector happens to fit into an integer type. But as the OP implies, the general form of the problem is really O(N).

An example of a solution to the general case is to use two pointers and a nested loop to simply compare every character to every other. The space requirement is O(1) but the time requirement is O(N^2).