Find the Maximum Length of Valid Subsequence II

For the known formulation, a subsequence is valid when every adjacent pair has the same sum modulo k. Let r be the current value modulo k and let p be the required pair-sum remainder. If a previous element has remainder prev=(p-r+k)%k, then appending the current value extends a sequence ending in prev whose established pair residue is p. Store dp[lastRemainder][pairRemainder] as the best length of a valid subsequence ending in that remainder; update it with max(existing, dp[prev][p]+1), with a single element as a base candidate. Iterate p across zero through k-1 for each number. Explain whether a one-element subsequence is considered valid, because it has no adjacent pair. Test repeated residues, k one, negative values after normalized modulo, and length one. The compact DP takes O(n*k) time and O(k squared) space, far better than enumerating O(2^n) subsequences.