Distinct Subsequences DP explanation

From LeetCode

Given a string S and a string T, count the number of distinct subsequences of T in S.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).

Here is an example: S = "rabbbit", T = "rabbit"

Return 3.

I see a very good DP solution, however, I have hard time to understand it, anybody can explain how this dp works?

int numDistinct(string S, string T) {

    vector<int> f(T.size()+1);

    //set the last size to 1.
    f[T.size()]=1;

    for(int i=S.size()-1; i>=0; --i){
        for(int j=0; j<T.size(); ++j){
            f[j]+=(S[i]==T[j])*f[j+1];
            printf("%dt", f[j] );
        }
        cout<<"n";
    }
    return f[0];
}

---

First, try to solve the problem yourself to come up with a naive implementation:

Let's say that S.length = m and T.length = n . Let's write S{i} for the substring of S starting at i . For example, if S = "abcde" , S{0} = "abcde" , S{4} = "e" , and S{5} = "" . We use a similar definition for T .

Let N[i][j] be the distinct subsequences for S{i} and T{j} . We are interested in N[0][0] (because those are both full strings).

There are two easy cases: N[i][n] for any i and N[m][j] for j<n . How many subsequences are there for "" in some string S ? Exactly 1. How many for some T in "" ? Only 0.

Now, given some arbitrary i and j , we need to find a recursive formula. There are two cases.

If S[i] != T[j] , we know that N[i][j] = N[i+1][j] (I hope you can verify this for yourself, I aim to explain the cryptic algorithm above in detail, not this naive version).

If S[i] = T[j] , we have a choice. We can either 'match' these characters and go on with the next characters of both S and T , or we can ignore the match (as in the case that S[i] != T[j] ). Since we have both choices, we need to add the counts there: N[i][j] = N[i+1][j] + N[i+1][j+1] .

---

In order to find N[0][0] using dynamic programming, we need to fill the N table. We first need to set the boundary of the table:

N[m][j] = 0, for 0 <= j < n
N[i][n] = 1, for 0 <= i <= m

Because of the dependencies in the recursive relation, we can fill the rest of the table looping i backwards and j forwards:

for (int i = m-1; i >= 0; i--) {
    for (int j = 0; j < n; j++) {
        if (S[i] == T[j]) {
            N[i][j] = N[i+1][j] + N[i+1][j+1];
        } else {
            N[i][j] = N[i+1][j];
        }
    }
}

---

We can now use the most important trick of the algorithm: we can use a 1-dimensional array f , with the invariant in the outer loop: f = N[i+1]; This is possible because of the way the table is filled. If we apply this to my algorithm, this gives:

f[j] = 0, for 0 <= j < n
f[n] = 1

for (int i = m-1; i >= 0; i--) {
    for (int j = 0; j < n; j++) {
        if (S[i] == T[j]) {
            f[j] = f[j] + f[j+1];
        } else {
            f[j] = f[j];
        }
    }
}

---

We're almost at the algorithm you gave. First of all, we don't need to initialize f[j] = 0 . Second, we don't need assignments of the type f[j] = f[j] .

Since this is C++ code, we can rewrite the snippet

if (S[i] == T[j]) {
    f[j] += f[j+1];
}

to

f[j] += (S[i] == T[j]) * f[j+1];

and that's all. This yields the algorithm:

f[n] = 1

for (int i = m-1; i >= 0; i--) {
    for (int j = 0; j < n; j++) {
        f[j] += (S[i] == T[j]) * f[j+1];
    }
}

---

I think the answer is wonderful, but something may be not correct.

I think we should iterate backwards over i and j . Then we change to array N to array f , we looping j forwards for not overlapping the result last got.

for (int i = m-1; i >= 0; i--) {
    for (int j = 0; j < n; j++) {
        if (S[i] == T[j]) {
            N[i][j] = N[i+1][j] + N[i+1][j+1];
        } else {
            N[i][j] = N[i+1][j];
        }
    }
}

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