Question

link

Find a polynomial-time algorithm that takes a string of length n, and a number k, output the number of distinct k-character subsequences.

For instance, input “food” and number k=2, output should be 4. There are four distinct 2-character subsequences of “food”: “fo”, “fd”, “oo”, and “od”.

Solution

Similar to [Question] Number of distinct sub-sequence, we solve this problem with DP. The dp equation is a bit difficult to write.

The idea come from comment from gareth_rees:

Let θ(S, k) be the number of distinct k-character subsequences in the string S of length n.

Clearly θ(S, k) = 1 if n = k or k = 0

and θ(S, k) = 0 if n < k.

Otherwise, choose 1 unique char from S, and deduct k by 1, then do the DP calculation with the remaining part of S.

Look at this example: 

θ("food", 2) = θ("ood", 1) + θ("od", 1) + θ("", 1)
= (θ("od", 0) + θ("", 0)) + (θ("d", 0) + θ("", 0)) + 0
= (1 + 1) + (1 + 1)
= 4

“food” is divided into 3 parts. First part we choose “f” to be the first char, thus the value is θ(“ood”, 1). Second part we choose “o”, and final part we choose “d”.

Note that when we choose a char, it must never have been chosen before. In case of “food”, we only choose ‘f’, ‘o’, ‘d’ once for each.

This is a very difficult DP question, but the explanation really makes the answer easier. Read my implementation below.

Code

public int countSubSeq(String input, int k) {
    // assuming all input chars are small letter
    return choose(input, 0, k);
}

private int choose(String input, int start, int numChar) {
    int charLeft = input.length() - start;
    if (charLeft == numChar || numChar == 0) {
        return 1;
    } else if (charLeft < numChar || numChar < 0) {
        return 0;
    }
    // now numChar is smaller than charLeft, and larger than 0
    // start to pick a char (which is at first appearance)
    int total = 0;
    HashSet<Character> chosen = new HashSet<Character>();
    while (start < input.length()) {
        char currentChar = input.charAt(start);
        if (!chosen.contains(currentChar)) {
            // pick the char pointer by 'start'
            total += choose(input, start + 1, numChar - 1);
            chosen.add(currentChar);
        }
        start++;
    }
    return total;
}