Find length of Longest Palindromic Subsequence (Recursion & DP)

The Memoization Technique is basically an extension to the recursive approach so that we can overcome the problem of calculating redundant cases and thus decrease time complexity.

We can see that in each recursive call only the value of "n" and "m" (length) changes, so we can store and reuse the result of a function(..n, m..) call using a "n * m" 2-D array.

The 2-D array will store a particular state (n, m) if we get it the first time.

Now, if we come across the same state (n, m) again, instead of calculating it in exponential complexity, we can directly return its result stored in the table in constant time.

The time complexity of this solution is (n * m).

In addition, O(n * m) auxiliary space was used by the table.

This Print length of Longest Palindromic Subsequence (LPS) is a variation of the Longest Common Subsequence (LCS).

The only extra thing we need to do is to obtain a reverse string "s2" of the given string "s1". This will act as the second string for LCS.

The idea is to generate all subsequences of both "s1" and "s2" and print the longest matching subsequence.

Start from the last elements (n-1, m-1) of both subsequences:

If s1[n-1] == s2[m-1], this character will be part of the longest subsequences, so add it to the current subsequence and check for the remaining (0...n-2, 0...m-2) characters.

If s1[n-1] == s2[m-1], then check recursively for (0...n-2, 0...m-1) and (0...n-1, 0...m-2) and return the longest subsequence of these two.

If any of the subsequences is exhausted (m ==0, n ==0) return "0". This is the base condition.

This solution is exponential O(2^(m+n)) in term of time complexity.

In addition, O(2^n) auxiliary stack space was used for the recursion stack.