Leetcode: Is Subsequence

Given a string s and a string t, check if s is subsequence of t.

You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is 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"

Return true.

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

Return false.

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?

Basic Solution: DP, O(mn) time, O(m) space, m is the size of s, n is the size of t

public class Solution {
    public boolean isSubsequence(String s, String t) {
        if (s.length() == 0) return true;
        boolean[] dp = new boolean[s.length()+1];
        dp[0] = true;
        for (int i=1; i<=t.length(); i++) {
            for (int j=s.length(); j>=0; j--) {
                if (dp[j] || t.charAt(i-1)==s.charAt(j-1) && dp[j-1])
                    dp[j] = true;
            }
            if (dp[s.length()]) return true;
        }
        return false;
    }
}

Greedy Solution: O(n) time, O(1) space

public class Solution {
    public boolean isSubsequence(String s, String t) {
        if (s.length() == 0) return true;
        int match = 0;
        for (int i=0; i<t.length(); i++) {
            if (t.charAt(i) == s.charAt(match)) {
                match++;
            }
            if (match == s.length()) return true;
        }
        return false;
    }
}

 Follow Up:

The best solution is to create a map for String t, key is char, value is the index of appearance in ascending order

    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;
    }