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 6Input: 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
// 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
# 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[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 |
Javascript
<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> |
Output
6
Time Complexity: O(N2)
Auxiliary Space: O(N)
Related Article: Longest Palindromic Subsequence
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