392. Is Subsequence

Two-pointer scan is the linear-time solution. Initialize i = 0 (pointer into s) and j = 0 (pointer into t). While both pointers are in range, if s[i] == t[j] increment i; always increment j. Return i == s.length. The DP framing for the follow-up: define subproblem dp[i][j] = whether s[0..i-1] is a subsequence of t[0..j-1]. The transition is dp[i][j] = dp[i-1][j-1] if s[i-1] == t[j-1] else dp[i][j] = dp[i][j-1]. Base case dp[0][j] = true (empty string is a subsequence of anything). For the indexed-by-character preprocessing variant, build next[t.length][26] where next[i][c] is the next index >= i in t containing character c — each query then runs in O(s.length). Time O(s.length + t.length), space O(1).