- 1
Two Strings, One Table
Two-sequence DP compares PREFIXES: dp[i][j] answers the question for the first i chars of s and the first j of t — match takes the diagonal + 1, mismatch combines the neighbors.
Multiple choice - 2
Is Subsequence
One greedy walk over t decides it: always advance the t pointer, advance the s pointer only on a match — then reframe the same question as a prefix table to bridge into LCS.
DebugMultiple choiceTrace - 3
Longest Common Subsequence
THE prefix table: dp[i][j] = LCS of the first i chars of a and first j of b — match takes the diagonal + 1, mismatch takes the better neighbor, and dp row i pairs with a.charAt(i-1), not a.charAt(i).
Complete codeDebugMultiple choiceTrace - 4
Longest Palindromic Subsequence
A palindrome reads the same in both directions, so it survives in reverse(s) too — LPS(s) collapses to LCS(s, reverse(s)), the table you already know.
Complete codeMultiple choiceTrace - 5
Word Break
dp[i] = can the first i characters be segmented — true when SOME split point j has dp[j] true and s[j..i) in the dictionary. Every split point stays in play; greedy longest-match does not.
Complete codeMultiple choiceTrace - 6
Delete Operation for Two Strings
Both strings can only shrink, so what survives is a common subsequence — keep the longest one and delete everything else: steps = m + n - 2·LCS.
Complete codeMultiple choiceTrace - 7
Sequence DP CheckpointCheckpoint
No new concepts — carry the prefix table to unseen ground: the shortest string that contains two others, crossing-free lines between arrays, deletions that charge by ASCII weight, and counting the ways a pattern hides inside a string.
Complete codeMultiple choiceTrace - 8
Sequence DP ReviewReview
Mixed recap of the island: what a prefix-table cell actually says, match-diagonal versus mismatch-neighbors, the reductions catalog, where the answer lives for ending-style versus prefix-style states, the segmentation loop, and rolling a table down to two rows.
Complete codeCode orderMultiple choice - 9
Boss: Longest Increasing SubsequenceBoss
The island's final test: the longest strictly increasing subsequence. One ending-style recurrence prices every chain at once — and the boss's traps are reading the last cell as the answer, settling for contiguous runs, and letting equal elements pretend to climb.
Complete codeDebugCode orderMultiple choiceTrace
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