Problem statement

Given two strings s and t, return true if s is a subsequence of t, or false otherwise.

A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not).

Example 1

Input: s = "abc", t = "ahbgdc"
Output: true

Example 2

Input: s = "axc", t = "ahbgdc"
Output: false

Constraints

• 0 <= s.length <= 100.
• 0 <= t.length <= 10^4.
• s and t consist only of lowercase English letters.

Follow up: Suppose there are lots of incoming s, say s1, s2, ..., sk where k >= 10^9, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code?

Solution

For each character of s, find if it exists in t. If not, return false.

Otherwise, continue with the next character of s and the finding starting from the position after the last found one.

Code

#include <iostream>
using namespace std;
bool isSubsequence(string s, string t) {
    int i = -1;
    for (char c : s) {
        i++;
        while (i < t.length() && t[i] != c) {
            i++;
        }
        if (i >= t.length()){
            return false;
        }
    }
    return true;
}
int main() {
    cout << isSubsequence("abc", "ahbgdc") << endl;
    cout << isSubsequence("axc", "ahbgdc") << endl;
    cout << isSubsequence("aaaa", "bbaaa") << endl;
}

Output:
1
0
0

Complexity

• Runtime: O(N), where N = t.length.
• Extra space: O(1).

Follow up

Suppose there are lots of incoming s, say s1, s2, ..., sk.

• If some is a subsequence of t then return true.
• If none of them satisfies, return false.

Code

#include <iostream>
#include <vector>
using namespace std;
bool isSubsequence(string s, string t) {
    int i = -1;
    for (char c : s) {
        i++;
        while (i < t.length() && t[i] != c) {
            i++;
        }
        if (i >= t.length()){
            return false;
        }
    }
    return true;
}
bool hasSubsequence(vector<string>& strings, string t) {
    for (string& s :  strings) {
        if (isSubsequence(s, t)) {
            return true;
        }
    }
    return false;
}
int main() {
    vector<string> strings{"abc", "axc", "aaaa"};
    cout << hasSubsequence(strings, "ahbgdc") << endl;
    strings = {"abcd", "axc", "aaaa"};
    cout << hasSubsequence(strings, "ahbgdc") << endl;
}

Output:
1
0

Complexity

• Runtime: O(kN), where k = strings.length and N = t.length.
• Extra space: O(1).