392 Is Subsequence

Given a stringsand a stringt, check ifsis subsequence oft.

You may assume that there is only lower case English letters in bothsandt.tis potentially a very long (length ~= 500,000) string, andsis a short string (<=100).

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).

Example 1: s="abc",t="ahbgdc"

Returntrue.

Example 2: s="axc",t="ahbgdc"

Returnfalse.

Follow up: If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

这题不知道为毛要放在DP里，原题的解决方法更像2pointer，follow up需要用到binary search来加速。

public boolean isSubsequence(String s, String t) {
    if (s == null || t == null) {
        return false;
    }

    int n = s.length();
    int m = t.length();

    int i = 0;
    int j = 0;

    while (i < n && j < m) {
        if (s.charAt(i) == t.charAt(j)) {
            i++;
            j++;
        } else {
            j++;
        }
    }

    if (i == n && j <= m) {
        return true;
    } else {
        return false;
    }
}

把discuss里大神写的follow up二分代码贴上来参考，自己是想不出来的了：

    // Follow-up: O(N) time for pre-processing, O(Mlog?) for each S.
    // Eg-1. s="abc", t="bahbgdca"
    // idx=[a={1,7}, b={0,3}, c={6}]
    //  i=0 ('a'): prev=1
    //  i=1 ('b'): prev=3
    //  i=2 ('c'): prev=6 (return true)
    // Eg-2. s="abc", t="bahgdcb"
    // idx=[a={1}, b={0,6}, c={5}]
    //  i=0 ('a'): prev=1
    //  i=1 ('b'): prev=6
    //  i=2 ('c'): prev=? (return false)
    public boolean isSubsequence(String s, String t) {
        List<Integer>[] idx = new List[256]; // Just for clarity
        for (int i = 0; i < t.length(); i++) {
            if (idx[t.charAt(i)] == null)
                idx[t.charAt(i)] = new ArrayList<>();
            idx[t.charAt(i)].add(i);
        }

        int prev = 0;
        for (int i = 0; i < s.length(); i++) {
            if (idx[s.charAt(i)] == null) return false; // Note: char of S does NOT exist in T causing NPE
            int j = Collections.binarySearch(idx[s.charAt(i)], prev);
            if (j < 0) j = -j - 1;
            if (j == idx[s.charAt(i)].size()) return false;
            prev = idx[s.charAt(i)].get(j) + 1;
        }
        return true;
    }