392. Is Subsequence

Easy

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

• s and t consist 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?

Intuition :

The problem requires checking if the characters in ( s ) appear in ( t ) in the same order. This can be achieved by iterating over ( t ) and trying to match each character in ( s ) sequentially. If all characters of ( s ) are found in ( t ) in the required order, then ( s ) is a subsequence of ( t ).

Approach :

1. Two Pointers: Use two pointers, one for ( s ) and one for ( t ).

2. Character Matching:

   • Traverse ( t ) using the ( t )-pointer.

   • Each time a character in ( t ) matches the current character in ( s ), move the ( s )-pointer to the next character.

3. Check Completion: If the ( s )-pointer reaches the end of ( s ), then all characters were matched in order, so return true.

4. End Condition: If the loop completes and the ( s )-pointer has not reached the end, return false.

Solving Steps ;

1. Initialize pointers for both ( s ) and ( t ) at the start.

2. Iterate over ( t ) and try to match each character in ( s ).

3. If the ( s )-pointer reaches the length of ( s ), return true.

4. If the iteration ends without completing ( s ), return false.

Code:

class Solution {
    public boolean isSubsequence(String s, String t) {

    int indexS = 0;
    int indexT = 0;
    while(indexS < s.length() && indexT < t.length()){

        if(t.charAt(indexT) == s.charAt(indexS))
        {
            indexS++;
        }
        indexT++;
      }

      return indexS == s.length();
    }
}