isSubsequence

• in this we will use 2 pointer where we will be putting both pointer at start of both the strings
• using loop iteration we will be comparing them char by char
• when they are equal we will increment both, if not we will increment t.
• in the end we will return if all are found .

Trigger: Checking relative order preservation across two strings. Mechanical Approach: Two Pointers.

• Pointer i on s, pointer j on t.
• If s[i] == t[j], move both. Else move j.
• Valid if i fully traverses s (i == len(s)).

The System/Scale Follow-up:

• Scenario: 1 massive string t, $10^9$ incoming short strings s.
• Architecture: Don’t rescan t. Precompute t into a Hash Map of char -> [list of indices].
• Execution: For each char in s, use Binary Search (bisect_right) on the Hash Map’s index list to find the next valid, strictly greater index in $O(\log N)$ time.

Blindspot: Always handle s = "" early. It should return True.