Length of longest Palindromic Subsequence of even length with no two adjacent characters same
Last Updated :
12 Jul, 2025
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:
Input: str = "abscrcdba"
Output: 6
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 <bits/stdc++.h>
using namespace std;
#define lli long long int
// To store the values of subproblems
unordered_map<string, lli> 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[0] == s[1] && s[0] == 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[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 = 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 implementation of above approach
import java.util.*;
class GFG {
// To store the values of subproblems
static Map<String, Integer> 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 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# 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[0] == s[1] && s[0] == 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[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) {
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
<script>
// Javascript implementation of above approach
// To store the values of subproblems
var dp = new Map();
// Function to find the
// Longest Palindromic subsequence of even
// length with no two adjacent characters same
function solve(s, 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.has(s + " " + c))
return dp.get(s + " " + c);
var 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 (var c1 = 'a'.charCodeAt(0); c1 <= 'z'.charCodeAt(0); c1++)
if (String.fromCharCode(c1) != c)
ans = Math.max(
ans,
1 + solve(s.substring(1,
s.length - 1),
String.fromCharCode(c1)));
}
// If it does not match
else {
// Then find the first and last index of
// given character in the given string
for (var i = 0; i < s.length; i++)
{
if (s[i] == c)
{
for (var 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.substring(i, j + 1),
c);
break;
}
break;
}
}
}
// Store the answer for future use
dp.set(s + " " + c, ans);
return ans;
}
// Driver code
var s = "abscrcdba";
var ma = 0;
// Check for all starting characters
for (var c1 = 'a'.charCodeAt(0); c1 <= 'z'.charCodeAt(0); c1++)
ma = Math.max(ma, solve(s, String.fromCharCode(c1)) * 2);
document.write(ma);
</script>
Time Complexity: O(N2)
Auxiliary Space: O(N)
Related Article: Longest Palindromic Subsequence
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