Is Subsequence

Description

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 sis 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?

Solutions

For original

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int i = 0;
        for(int j = 0; i < s.size() && j < t.size(); ++j){
            if(s[i] == t[j]){
                ++i;
            }
        }
        return i == s.size();
    }
};

For follow-up

Store the indices of characters in t in a hash table T.

For each s, use the following algorithm check is s is a subsequence of t.

Let i be the index of the first unused character in t. Initially i = 0.

For each character c in s, find the first index of c in t[i..]. The indices of c are stored in T[c] as a sorted list. We can use binary search to find the index in O(t.size()) time. If no such index, return false. Otherwise set i to the index + 1.

Let n be the average length of s, k be the number of s, m be the length of t. Then the running time is O(m + k * n * log(m)). running time using the solution for the original problem is O(k * m).

class Solution {
public:
    bool isSubsequence(string s, string t) {
        vector<vector<int>> hash(26);
        for(int i = 0; i < t.size(); ++i){
            hash[t[i] - 'a'].push_back(i);
        }
        return isSubsequence(s, hash);
    }
    
    bool isSubsequence(string &s, vector<vector<int>> &t){
        int i = 0;
        for(char c: s){
            auto indices = t[c - 'a'];
            auto it = lower_bound(indices.begin(), indices.end(), i);
            if(it == indices.end())
                return false;
            else
                i = *it + 1;
        }
        return true;
    }
};