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# Length of longest Palindromic Subsequence of even length with no two adjacent characters same

• Last Updated : 08 Oct, 2021

Given a string str, the task is to find the length of the longest palindromic subsequence of even length with no two adjacent characters same except the middle characters.

Examples:

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Input: str = “abscrcdba”
Output:
Explanation:
abccba is the required string which has no two consecutive characters same except the middle characters. Hence the length is 6

Input: str = “abcd”
Output: 0

Approach: The idea is to form a recursive solution and store the values of the subproblems using Dynamic Programming. The following steps can be followed to compute the result:

• Form a recursive function which will take a string and a character which is the starting character of the subsequence.
• If the first and last character of the string matches with the given character then remove the first and last character and call the function with all the character values from ‘a’ to ‘z’ except the given character, as the adjacent character cannot be same and find the maximum length.
• If the first and last character of the string does not match with the given character, then find the first and last index of the given character in the string, say i, j respectively. Take the substring from i to j and call the function with substring and the given character.
• Finally, memorise the values in an unordered map and use it if the function is again called with the same parameters.

The following is the implementation of the above approach:

## C++

 `// C++ implementation of above approach`` ` `#include ``using` `namespace` `std;`` ` `#define lli long long int`` ` `// To store the values of subproblems``unordered_map dp;`` ` `// Function to find the``// Longest Palindromic subsequence of even``// length with no two adjacent characters same``lli solve(string s, ``char` `c)``{``    ``// Base cases``    ``// If the string length is 1 return 0``    ``if` `(s.length() == 1)``        ``return` `0;`` ` `    ``// If the string length is 2``    ``if` `(s.length() == 2) {`` ` `        ``// Check if the characters match``        ``if` `(s == s && s == c)``            ``return` `1;``        ``else``            ``return` `0;``    ``}`` ` `    ``// If the value with given parameters is``    ``// previously calculated``    ``if` `(dp[s + ``" "` `+ c])``        ``return` `dp[s + ``" "` `+ c];`` ` `    ``lli ans = 0;`` ` `    ``// If the first and last character of the``    ``// string matches with the given character``    ``if` `(s == s[s.length() - 1] && s == c) ``    ``{``        ``// Remove the first and last character``        ``// and call the function for all characters``        ``for` `(``char` `c1 = ``'a'``; c1 <= ``'z'``; c1++)``            ``if` `(c1 != c)``                ``ans = max(``                    ``ans,``                    ``1 + solve(s.substr(1, ``                                       ``s.length() - 2),``                                       ``c1));``    ``}`` ` `    ``// If it does not match``    ``else` `{`` ` `        ``// Then find the first and last index of``        ``// given character in the given string``        ``for` `(lli i = 0; i < s.length(); i++) ``        ``{``            ``if` `(s[i] == c) ``            ``{``                ``for` `(lli j = s.length() - 1; j > i; j--)``                    ``if` `(s[j] == c) ``                    ``{``                        ``if` `(j == i)``                            ``break``;`` ` `                        ``// Take the substring from i``                        ``// to j and call the function``                        ``// with substring``                        ``// and the given character``                        ``ans = solve(s.substr(i, j - i + 1),``                                    ``c);``                        ``break``;``                    ``}`` ` `                ``break``;``            ``}``        ``}``    ``}`` ` `    ``// Store the answer for future use``    ``dp[s + ``" "` `+ c] = ans;``    ``return` `dp[s + ``" "` `+ c];``}`` ` `// Driver code``int` `main()``{``    ``string s = ``"abscrcdba"``;`` ` `    ``lli ma = 0;`` ` `    ``// Check for all starting characters``    ``for` `(``char` `c1 = ``'a'``; c1 <= ``'z'``; c1++)``        ``ma = max(ma, solve(s, c1) * 2);``    ``cout << ma << endl;`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of above approach``import` `java.util.*;`` ` `class` `GFG {`` ` `    ``// To store the values of subproblems``    ``static` `Map dp = ``new` `HashMap<>();`` ` `    ``// Function to find the``    ``// Longest Palindromic subsequence of even``    ``// length with no two adjacent characters same``    ``static` `Integer solve(``char``[] s, ``char` `c)``    ``{`` ` `        ``// Base cases``        ``// If the String length is 1 return 0``        ``if` `(s.length == ``1``)``            ``return` `0``;`` ` `        ``// If the String length is 2``        ``if` `(s.length == ``2``) {`` ` `            ``// Check if the characters match``            ``if` `(s[``0``] == s[``1``] && s[``0``] == c)``                ``return` `1``;``            ``else``                ``return` `0``;``        ``}`` ` `        ``// If the value with given parameters is``        ``// previously calculated``        ``if` `(dp.containsKey(String.valueOf(s) + ``" "` `+ c))``            ``return` `dp.get(String.valueOf(s) + ``" "` `+ c);``        ``Integer ans = ``0``;`` ` `        ``// If the first and last character of the``        ``// String matches with the given character``        ``if` `(s[``0``] == s[s.length - ``1``] && s[``0``] == c)``        ``{``            ``// Remove the first and last character``            ``// and call the function for all characters``            ``for` `(``char` `c1 = ``'a'``; c1 <= ``'z'``; c1++)`` ` `                ``if` `(c1 != c)``                    ``ans = Math.max(``                        ``ans,``                        ``1` `+ solve(Arrays.copyOfRange(``                                        ``s, ``1``, ``                                        ``s.length - ``1``),``                                        ``c1));``        ``}`` ` `        ``// If it does not match``        ``else` `{`` ` `            ``// Then find the first and last index of``            ``// given character in the given String``            ``for` `(Integer i = ``0``; i < s.length; i++) ``            ``{``                ``if` `(s[i] == c) ``                ``{``                    ``for` `(Integer j = s.length - ``1``; ``                         ``j > i;``                         ``j--)``                        ``if` `(s[j] == c) ``                        ``{``                            ``if` `(j == i)``                                ``break``;`` ` `                            ``// Take the subString from i``                            ``// to j and call the function``                            ``// with subString``                            ``// and the given character``                            ``ans = solve(Arrays.copyOfRange(``                                            ``s, i, ``                                            ``j + ``1``),``                                            ``c);``                            ``break``;``                        ``}`` ` `                    ``break``;``                ``}``            ``}``        ``}`` ` `        ``// Store the answer for future use``        ``dp.put(String.valueOf(s) + ``" "` `+ c, ans);``        ``return` `dp.get(String.valueOf(s) + ``" "` `+ c);``    ``}`` ` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String s = ``"abscrcdba"``;`` ` `        ``Integer ma = ``0``;`` ` `        ``// Check for all starting characters``        ``for` `(``char` `c1 = ``'a'``; c1 <= ``'z'``; c1++)``            ``ma = Math.max(ma,``                          ``solve(s.toCharArray(), c1) * ``2``);``        ``System.out.print(ma + ``"\n"``);``    ``}``}`` ` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of above approach`` ` `# To store the values of subproblems``dp ``=` `{}`` ` `# Function to find the``# Longest Palindromic subsequence of even``# length with no two adjacent characters same`` ` ` ` `def` `solve(s, c):`` ` `    ``# Base cases``    ``# If the string length is 1 return 0``    ``if` `(``len``(s) ``=``=` `1``):``        ``return` `0`` ` `    ``# If the string length is 2``    ``if` `(``len``(s) ``=``=` `2``):`` ` `        ``# Check if the characters match``        ``if` `(s[``0``] ``=``=` `s[``1``] ``and` `s[``0``] ``=``=` `c):``            ``return` `1``        ``else``:``            ``return` `0`` ` `    ``# If the value with given parameters is``    ``# previously calculated``    ``if` `(s ``+` `" "` `+` `c) ``in` `dp:``        ``return` `dp[s ``+` `" "` `+` `c]`` ` `    ``ans ``=` `0`` ` `    ``# If the first and last character of the``    ``# string matches with the given character``    ``if` `(s[``0``] ``=``=` `s[``len``(s) ``-` `1``] ``and` `s[``0``] ``=``=` `c):`` ` `        ``# Remove the first and last character``        ``# and call the function for all characters``        ``for` `c1 ``in` `range``(``97``, ``123``):`` ` `            ``if` `(``chr``(c1) !``=` `c):``                ``ans ``=` `max``(ans, ``1` `+` `solve(``                    ``s[``1``: ``len``(s) ``-` `1``], ``chr``(c1)))`` ` `    ``# If it does not match``    ``else``:`` ` `        ``# Then find the first and last index of``        ``# given character in the given string``        ``for` `i ``in` `range``(``len``(s)):`` ` `            ``if` `(s[i] ``=``=` `c):``                ``for` `j ``in` `range``(``len``(s) ``-` `1``, i, ``-``1``):``                    ``if` `(s[j] ``=``=` `c):``                        ``if` `(j ``=``=` `i):``                            ``break`` ` `                        ``# Take the substring from i``                        ``# to j and call the function``                        ``# with substring``                        ``# and the given character``                        ``ans ``=` `solve(s[i: j ``-` `i ``+` `2``], c)``                        ``break`` ` `                ``break`` ` `    ``# Store the answer for future use``    ``dp[s ``+` `" "` `+` `c] ``=` `ans`` ` `    ``return` `dp[s ``+` `" "` `+` `c]`` ` ` ` `# Driver code``if` `__name__ ``=``=` `"__main__"``:`` ` `    ``s ``=` `"abscrcdba"`` ` `    ``ma ``=` `0`` ` `    ``# Check for all starting characters``    ``for` `c1 ``in` `range``(``97``, ``123``):``        ``ma ``=` `max``(ma, solve(s, ``chr``(c1)) ``*` `2``)`` ` `    ``print``(ma)`` ` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of``// the above approach``using` `System;``using` `System.Collections;``using` `System.Collections.Generic;`` ` `class` `GFG {`` ` `    ``// To store the values of subproblems``    ``static` `Dictionary<``string``, ``int``> dp``        ``= ``new` `Dictionary<``string``, ``int``>();`` ` `    ``// Function to find the``    ``// Longest Palindromic subsequence of even``    ``// length with no two adjacent characters same``    ``static` `int` `solve(``char``[] s, ``char` `c)``    ``{`` ` `        ``// Base cases``        ``// If the String length is 1 return 0``        ``if` `(s.Length == 1)``            ``return` `0;`` ` `        ``// If the String length is 2``        ``if` `(s.Length == 2) {`` ` `            ``// Check if the characters match``            ``if` `(s == s && s == c)``                ``return` `1;``            ``else``                ``return` `0;``        ``}`` ` `        ``// If the value with given parameters is``        ``// previously calculated``        ``if` `(dp.ContainsKey(``new` `string``(s) + ``" "` `+ c))``            ``return` `dp[``new` `string``(s) + ``" "` `+ c];`` ` `        ``int` `ans = 0;`` ` `        ``// If the first and last character of the``        ``// String matches with the given character``        ``if` `(s == s[s.Length - 1] && s == c) ``        ``{`` ` `            ``// Remove the first and last character``            ``// and call the function for all characters``            ``for` `(``char` `c1 = ``'a'``; c1 <= ``'z'``; c1++)`` ` `                ``if` `(c1 != c) {``                    ``int` `len = s.Length - 2;``                    ``char``[] tmp = ``new` `char``[len];`` ` `                    ``Array.Copy(s, 1, tmp, 0, len);``                    ``ans = Math.Max(ans, 1 + solve(tmp, c1));``                ``}``        ``}`` ` `        ``// If it does not match``        ``else` `{`` ` `            ``// Then find the first and last index of``            ``// given character in the given String``            ``for` `(``int` `i = 0; i < s.Length; i++) ``            ``{``                ``if` `(s[i] == c) ``                ``{``                    ``for` `(``int` `j = s.Length - 1; j > i; j--)``                        ``if` `(s[j] == c) ``                        ``{``                            ``if` `(j == i)``                                ``break``;`` ` `                            ``// Take the subString from i``                            ``// to j and call the function``                            ``// with subString and the``                            ``// given character``                            ``int` `len = j + 1 - i;``                            ``char``[] tmp = ``new` `char``[len];``                            ``Array.Copy(s, i, tmp, 0, len);``                            ``ans = solve(tmp, c);``                            ``break``;``                        ``}``                    ``break``;``                ``}``            ``}``        ``}`` ` `        ``// Store the answer for future use``        ``dp[``new` `string``(s) + ``" "` `+ c] = ans;``        ``return` `dp[``new` `string``(s) + ``" "` `+ c];``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``string` `s = ``"abscrcdba"``;`` ` `        ``int` `ma = 0;`` ` `        ``// Check for all starting characters``        ``for` `(``char` `c1 = ``'a'``; c1 <= ``'z'``; c1++)``            ``ma = Math.Max(ma,``                          ``solve(s.ToCharArray(), c1) * 2);`` ` `        ``Console.Write(ma + ``"\n"``);``    ``}``}`` ` `// This code is contributed by rutvik_56`

## Javascript

 ``
Output
`6`

Time Complexity: O(N2)
Auxiliary Space: O(N)

Related Article: Longest Palindromic Subsequence

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