Solution

The correct answer is Option 3) 2,3,2,1

Key Points

• LCS (Longest Common Subsequence) is a sequence that appears in both lists in the same order (not necessarily contiguously) and has maximum possible length.
• Let A = {1,2,3,2,4,1,2} and B = {2,4,3,1,2,1}. We will use the standard DP (dynamic programming) table of lengths L[i][j] = LCS length of A[1..i] and B[1..j].

DP Length Table (final values)

i\j	1(2)	2(4)	3(3)	4(1)	5(2)	6(1)
1(1)	0	0	0	1	1	1
2(2)	1	1	1	1	2	2
3(3)	1	1	2	2	2	2
4(2)	1	1	2	2	3	3
5(4)	1	2	2	2	3	3
6(1)	1	2	2	3	3	4
7(2)	1	2	2	3	4	4

The bottom-right value is 4, so the LCS length is 4.

Reconstruction (backtracking idea)

• From the last cell (length 4), match the last common element: A[6]=1 matches B[6]=1 → include 1.
• Move to subproblem before these indices: we are now at length 3 with prefixes ending before A[6] and B[6].
• Trace previous matches: before the final 1, we can match 2 (A[4] with B[5]), before that 3 (A[3] with B[3]), and before that 2 (A[2] with B[1]).

Thus one LCS is {2, 3, 2, 1}.

Why other options are wrong

• 2,1,2,3: In B, after taking the last 2 (position 5), there is no 3 to the right, so this order cannot occur.
• 1,3,2,1: In B, 1 appears at positions 4 and 6; there is no 3 after position 4 that can still keep the order 1→3, so this sequence cannot occur.
• 2,3,1,2,1 (length 5): Not a subsequence of A (after 1 at position 6, there is no trailing 1 to complete length 5); hence impossible, and anyway longer than the DP length 4.

Conclusion: The longest common subsequence of the two sequences is 2,3,2,1 (length 4).