Longest Palindromic Subsequence in Python Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Longest Palindromic Subsequence (LPS) problem is about finding the longest subsequence of the given sequence which is a palindrome. In Python, the task of maximizing the number of words in a sentence can be solved by the methods of dynamic programming. The algorithm for finding the Longest Palindromic Subsequence involves creating a table to store the lengths of the longest palindromic subsequences for different substrings of the input sequence. Longest Palindromic Subsequence in Python using Memoization:Step-by-step algorithm: Define a recursive function, say lps(s1, s2, n1, n2) which finds the longest common subsequence among s1 and s2 where n1 is index of s1 and n2 is index of s2.Call the lps function with input string seq as s1 and reverse of seq as s2.If we reach the first index of any string, return 0Check if we have already computed the result for the current indices.If the result is already computed, return the result.Otherwise, if the current indices of both s1 and s2 are equal, dp[n1][n2] = 1 + lps(s1, s2, n1 - 1, n2 - 1)Else if the current index of both s1 and s2 are not equal, dp[n1][n2] = max(lps(s1, s2, n1 - 1, n2), lps(s1, s2, n1, n2 - 1))Store the above result in dp[n1][n2]Below is the implementation of Longest Palindromic Subsequence in Python using Memoization: Python # A Dynamic Programming based Python program for LPS problem # Returns the length of the longest palindromic subsequence # in seq dp = [[-1 for i in range(1001)]for j in range(1001)] # Returns the length of the longest palindromic subsequence # in seq def lps(s1, s2, n1, n2): if (n1 == 0 or n2 == 0): return 0 if (dp[n1][n2] != -1): return dp[n1][n2] if (s1[n1 - 1] == s2[n2 - 1]): dp[n1][n2] = 1 + lps(s1, s2, n1 - 1, n2 - 1) return dp[n1][n2] else: dp[n1][n2] = max(lps(s1, s2, n1 - 1, n2), lps(s1, s2, n1, n2 - 1)) return dp[n1][n2] # Driver program to test above functions seq = "GEEKSFORGEEKS" n = len(seq) s2 = seq s2 = s2[::-1] print(f"The length of the LPS is {lps(s2, seq, n, n)}") OutputThe length of the LPS is 5 Time Complexity: O(n2), where n is the length of the input stringAuxiliary Space: O(n2) Longest Palindromic Subsequence in Python using Tabulation:Step-by-step algorithm: Initialize a 2D array dp with dimensions (n x n), where n is the length of the input sequence.Set the diagonal elements of dp to 1, as each single character is a palindrome of length 1.Traverse the input sequence with two pointers i and j, where i starts from n-1 and decrements, and j starts from i+1 and increments.If seq[i] is equal to seq[j], set dp[i][j] to dp[i+1][j-1] + 2 because the current characters form a palindromic subsequence of length 2 greater than the one inside them.Otherwise, set dp[i][j] to the maximum of dp[i+1][j] and dp[i][j-1], as we discard one character from either end and try to find the longest palindromic subsequence.The result is stored in dp[0][n-1], which represents the length of the Longest Palindromic Subsequence.Below is the implementation of Longest Palindromic Subsequence in Python using Tabulation: Python def longestPalinSubseq(S): R = S[::-1] # dp[i][j] will store the length of the longest # palindromic subsequence for the substring # starting at index i and ending at index j dp = [[0] * (len(R) + 1) for _ in range(len(S) + 1)] # Filling up DP table based on conditions discussed # in the above approach for i in range(1, len(S) + 1): for j in range(1, len(R) + 1): if S[i - 1] == R[j - 1]: dp[i][j] = 1 + dp[i - 1][j - 1] else: dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]) # At the end, DP table will contain the LPS # So just return the length of LPS return dp[len(S)][len(R)] # Driver code s = "GEEKSFORGEEKS" print("The length of the LPS is", longestPalinSubseq(s)) OutputThe length of the LPS is 5 Time Complexity: O(n2), where n is the length of the input stringAuxiliary Space: O(n2) Comment More infoAdvertise with us Next Article Asymptotic Notations for Analysis of Algorithms A amit7822i5n Follow Improve Article Tags : Strings Dynamic Programming DSA Python-DSA Practice Tags : Dynamic ProgrammingStrings Similar Reads Basics & PrerequisitesTime and Space ComplexityMany times there are more than one ways to solve a problem with different algorithms and we need a way to compare multiple ways. 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