Is Subsequence
Problem Description
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 <= 104
sandtconsist only of lowercase English letters.
Follow up: Suppose there are lots of incoming s, say s1, s2, ..., sk where k >= 109, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code?
Solve it on LeetCode
Approaches
1. Two Pointers
Intuition:
To check whether a string s is a subsequence of t, we try to match the characters of s in order while scanning through t. We don’t need to modify either string or use extra space.
We use two pointers:
- one pointer moves through s
- the other moves through t
As we traverse t, we look for characters that match the current character in s. Whenever we find a match, we advance the pointer in s.
The pointer in t always moves because we are scanning the entire sequence.
If we manage to advance through all characters in s, then every character in s appears in the correct order in t, meaning s is a subsequence of t.
Steps:
- Initialize two pointers,
ifor tracking the current character insandjfor scanning throught - Iterate over
tusingj. - Whenever
s[i] == t[j], incrementibecause we matched a character ofs. - Always increment
jbecause we continue scanningt. - If
ireaches the length ofs, it means all characters inshave been matched correctly in order withint. - Return
trueif all characters ofswere matched; otherwise returnfalse.
Code:
- Time Complexity: O(n), where n is the length of the array.
- Space Complexity: O(n), due to the use of additional array.