Given a string, the task is to return the length of its longest possible chunked palindrome. It means a palindrome formed by substring in the case when it is not formed by characters of the string. For a better understanding look at the example
Examples:
Input : ghiabcdefhelloadamhelloabcdefghi Output : 7 (ghi)(abcdef)(hello)(adam)(hello)(abcdef)(ghi) Input : merchant Output : 1 (merchant) Input : antaprezatepzapreanta Output : 11 (a)(nt)(a)(pre)(za)(tpe)(za)(pre)(a)(nt)(a) Input : neveropen Output : 3 (neveropen)(for)(neveropen)
The entire idea is to create chunks from left and right and recursively.
As you can see, we can match substring from the left side chunk and match it with the exact right side chunk. Once we get a match, we recursively count the length of the longest possible chunked palindrome in the remaining string. We end the recursion once no string is left or when no more valid chunked parts can be found.
Implementation:
C++
// C++ program to find the length of longest palindromic chunk#include <bits/stdc++.h>using namespace std;// Here curr_str is the string whose LCP is needed// len is length of string evaluated till now// and str is original stringint LPCRec(string curr_str, int count, int len, string str) { // if there is noting at all!! if (curr_str.size()==0) return (0); // if a single letter is left out if (curr_str.size() <= 1) { if (count != 0 && (str.size() - len )<= 1) return (count + 1); else return (1); } // for each length of substring chunk in string int n = curr_str.size(); for (int i = 0; i < n/2; i++) { // if left side chunk and right side chunk // are same if (curr_str.substr(0, i+1) == curr_str.substr(n-1-i,i+1)) { // Call LCP for the part between the // chunks and add 2 to the result. // Length of string evaluated till // now is increased by (i+1)*2 return LPCRec(curr_str.substr(i+1, n-2-2*i), // count + 2, len + (i+1)*2, str); } } return count + 1; }// Wrapper over LPCRec()int LPC(string str){ return LPCRec(str, 0, 0, str);}// driver functionint main(){ cout<<"V : "<<LPC("V")<<endl; cout<<"VOLVO : " << LPC("VOLVO")<<endl; cout<<"VOLVOV : " << LPC("VOLVOV")<<endl; cout<<"ghiabcdefhelloadamhelloabcdefghi : "<< LPC("ghiabcdefhelloadamhelloabcdefghi")<<endl; cout<<"ghiabcdefhelloadamhelloabcdefghik : "<< LPC("ghiabcdefhelloadamhelloabcdefghik")<<endl; cout<<"antaprezatepzapreanta : " << LPC("antaprezatepzapreanta");}// This code is contributed by Pushpesh Raj |
Java
/* Java program to find the length of longest palindromic chunk */import java.util.*;import java.lang.*;import java.io.*;class LongestPalindromicChunk{ // Here s is the string whose LCP is needed // ln is length of string evaluated till now // and str is original string private static int LPCRec(String curr_str, int count, int len, String str) { // if there is noting at all!! if (curr_str == null || curr_str.isEmpty()) return (0); // if a single letter is left out if (curr_str.length() <= 1) { if (count != 0 && str.length() - len <= 1) return (count + 1); else return (1); } // for each length of substring chunk in string int n = curr_str.length(); for (int i = 0; i < n/2; i++) { // if left side chunk and right side chunk // are same if (curr_str.substring(0, i + 1). equals(curr_str.substring(n-1-i, n))) { // Call LCP for the part between the // chunks and add 2 to the result. // Length of string evaluated till // now is increased by (i+1)*2 return LPCRec(curr_str.substring(i+1, n-1-i), count + 2, len + (i+1)*2, str); } } return count + 1; } // Wrapper over LPCRec() public static int LPC(String str) { return LPCRec(str, 0, 0, str); } // driver function public static void main(String[] args) { System.out.println("V : " + LPC("V")); System.out.println("VOLVO : " + LPC("VOLVO")); System.out.println("VOLVOV : " + LPC("VOLVOV")); System.out.println("ghiabcdefhelloadamhelloabcdefghi : " + LPC("ghiabcdefhelloadamhelloabcdefghi")); System.out.println("ghiabcdefhelloadamhelloabcdefghik : " + LPC("ghiabcdefhelloadamhelloabcdefghik")); System.out.println("antaprezatepzapreanta : " + LPC("antaprezatepzapreanta")); }} |
Python3
# Python3 program to find length of # longest palindromic chunk# Here curr_str is the string whose # LCP is needed leng is length of # string evaluated till now and s # is original string def LPCRec(curr_str, count, leng, s): # If there is nothing at all!! if not curr_str: return 0 # If a single letter is left out if len(curr_str) <= 1: if count != 0 and len(s) - leng <= 1: return (count + 1) else: return 1 # For each length of substring # chunk in string n = len(curr_str) for i in range(n // 2): # If left side chunk and right # side chunk are same if (curr_str[0 : i + 1] == curr_str[n - 1 - i : n]): # Call LCP for the part between the # chunks and add 2 to the result. # Length of string evaluated till # now is increased by (i+1)*2 return LPCRec(curr_str[i + 1 : n - 1 - i], count + 2, leng + (i + 1) * 2, s) return count + 1# Wrapper over LPCRec() def LPC(s): return LPCRec(s, 0, 0, s)# Driver code print("V :", LPC("V"))print("VOLVO :", LPC("VOLVO"))print("VOLVOV :", LPC("VOLVOV"))print("ghiabcdefhelloadamhelloabcdefghi :", LPC("ghiabcdefhelloadamhelloabcdefghi"))print("ghiabcdefhelloadamhelloabcdefghik :", LPC("ghiabcdefhelloadamhelloabcdefghik"))print("antaprezatepzapreanta :", LPC("antaprezatepzapreanta"))# This code is contributed by Prateek Gupta |
C#
// C# program to find length of // longest palindromic chunkusing System;class GFG{// Here s is the string whose LCP // is needed ln is length of string // evaluated till now and str is // original string private static int LPCRec(string curr_str, int count, int len, string str){ // if there is noting at all!! if (string.ReferenceEquals(curr_str, null) || curr_str.Length == 0) { return (0); } // if a single letter is left out if (curr_str.Length <= 1) { if (count != 0 && str.Length - len <= 1) { return (count + 1); } else { return (1); } } // for each length of substring // chunk in string int n = curr_str.Length; for (int i = 0; i < n / 2; i++) { // if left side chunk and right side chunk // are same if (curr_str.Substring(0, i + 1).Equals( curr_str.Substring(n - 1 - i, n - (n - 1 - i)))) { // Call LCP for the part between the // chunks and add 2 to the result. // Length of string evaluated till // now is increased by (i+1)*2 return LPCRec(curr_str.Substring(i + 1, (n - 1 - i) - (i + 1)), count + 2, len + (i + 1) * 2, str); } } return count + 1;}// Wrapper over LPCRec() public static int LPC(string str){ return LPCRec(str, 0, 0, str);}// Driver Codepublic static void Main(string[] args){ Console.WriteLine("V : " + LPC("V")); Console.WriteLine("VOLVO : " + LPC("VOLVO")); Console.WriteLine("VOLVOV : " + LPC("VOLVOV")); Console.WriteLine("ghiabcdefhelloadamhelloabcdefghi : " + LPC("ghiabcdefhelloadamhelloabcdefghi")); Console.WriteLine("ghiabcdefhelloadamhelloabcdefghik : " + LPC("ghiabcdefhelloadamhelloabcdefghik")); Console.WriteLine("antaprezatepzapreanta : " + LPC("antaprezatepzapreanta"));}}// This code is contributed by Shrikant13 |
Javascript
<script>// JavaScript program to find length of // longest palindromic chunk// Here curr_str is the string whose // LCP is needed leng is length of // string evaluated till now and s // is original string function LPCRec(curr_str, count, leng, s){ // If there is nothing at all!! if(!curr_str) return 0 // If a single letter is left out if(curr_str.length <= 1){ if (count != 0 && s.length - leng <= 1) return (count + 1) else return 1 } // For each length of substring // chunk in string let n = curr_str.length for(let i=0;i<Math.floor(n/2);i++){ // If left side chunk and right // side chunk are same if (curr_str.substring(0,i+1) == curr_str.substring(n-1-i,n)) // Call LCP for the part between the // chunks and add 2 to the result. // Length of string evaluated till // now is increased by (i+1)*2 return LPCRec(curr_str.substring(i + 1,n - 1 - i), count + 2, leng + (i + 1) * 2, s) } return count + 1}// Wrapper over LPCRec() function LPC(s){ return LPCRec(s, 0, 0, s)}// Driver code document.write("V :", LPC("V"),"</br>")document.write("VOLVO :", LPC("VOLVO"),"</br>")document.write("VOLVOV :", LPC("VOLVOV"),"</br>")document.write("ghiabcdefhelloadamhelloabcdefghi :", LPC("ghiabcdefhelloadamhelloabcdefghi"),"</br>")document.write("ghiabcdefhelloadamhelloabcdefghik :", LPC("ghiabcdefhelloadamhelloabcdefghik"),"</br>")document.write("antaprezatepzapreanta :", LPC("antaprezatepzapreanta"),"</br>")// This code is contributed by shinjanpatra</script> |
Output
V : 1 VOLVO : 3 VOLVOV : 5 ghiabcdefhelloadamhelloabcdefghi : 7 ghiabcdefhelloadamhelloabcdefghik : 1 antaprezatepzapreanta : 11
Following is the C++ implementation with memoization for above problem.
Implementation:
CPP
#include <climits>#include <iostream>#include <unordered_map>using namespace std;unordered_map<string, int> mem;int process(string& s, int l, int r){ int ans = 1; if (l > r) return 0; // check if we've already solved this if (mem.find(s.substr(l, r - l + 1)) != mem.end()) return mem[s.substr(l, r - l + 1)]; for (int len = 1; len <= (r - l + 1) / 2; len++) { if (s.substr(l, len) == s.substr(r - len + 1, len)) ans = max(ans, 2 + process(s, l + len, r - len)); } // remember result for future mem[s.substr(l, r - l + 1)] = ans; return ans;}int LPC(string s) { return process(s, 0, s.length() - 1); }int main(){ cout << LPC("aaaaabaababbaabaaababababababababbbbaaaaa") << endl; return 0;} |
Java
import java.util.HashMap;public class Main { static HashMap<String, Integer> mem = new HashMap<>(); public static int process(String s, int l, int r) { int ans = 1; if (l > r) { return 0; } // check if we've already solved this if (mem.containsKey(s.substring(l, r + 1))) { return mem.get(s.substring(l, r + 1)); } for (int len = 1; len <= (r - l + 1) / 2; len++) { if (s.substring(l, l + len).equals(s.substring(r - len + 1, r + 1))) { ans = Math.max(ans, 2 + process(s, l + len, r - len)); } } // remember result for future mem.put(s.substring(l, r + 1), ans); return ans; } public static int LPC(String s) { return process(s, 0, s.length() - 1); } public static void main(String[] args) { System.out.println(LPC("aaaaabaababbaabaaababababababababbbbaaaaa")); }}// This code is contributed by Aman Kumar |
Python3
# Python code for the approachmem = {}def process(s,l,r): global mem ans = 1 if (l > r): return 0 # check if we've already solved this if (s[l: r+1] in mem): return mem[s[l: r+1]] for Len in range(1,(r-l+1)//2 + 1,1): if (s[l: Len+l] == s[r-Len+1:r+1]): ans = max(ans, 2 + process(s, l+Len, r-Len)) # remember result for future mem[s[l: r+1]] = ans return ansdef LPC(s): return process(s, 0, len(s)-1)# driver codeprint(LPC("aaaaabaababbaabaaababababababababbbbaaaaa"))# This code is contributed by shinjanpatra. |
Javascript
let mem = {};function process(s, l, r) { let ans = 1; if (l > r) { return 0; } // check if we've already solved this if (s.slice(l, r + 1) in mem) { return mem[s.slice(l, r + 1)]; } for (let Len = 1; Len <= Math.floor((r - l + 1) / 2); Len++) { if (s.slice(l, Len + l) === s.slice(r - Len + 1, r + 1)) { ans = Math.max(ans, 2 + process(s, l + Len, r - Len)); } } // remember result for future mem[s.slice(l, r + 1)] = ans; return ans;}function LPC(s) { return process(s, 0, s.length - 1);}// driver codeconsole.log(LPC("aaaaabaababbaabaaababababababababbbbaaaaa"));// This code is contributed by adityashatmfh |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class Program{ static Dictionary<string, int> mem = new Dictionary<string, int>(); static int Process(string s, int l, int r) { int ans = 1; if (l > r) return 0; // check if we've already solved this if (mem.ContainsKey(s.Substring(l, r - l + 1))) return mem[s.Substring(l, r - l + 1)]; for (int len = 1; len <= (r - l + 1) / 2; len++) { if (s.Substring(l, len) == s.Substring(r - len + 1, len)) ans = Math.Max(ans, 2 + Process(s, l + len, r - len)); } // remember result for future mem[s.Substring(l, r - l + 1)] = ans; return ans; } static int LPC(string s) { return Process(s, 0, s.Length - 1); } static void Main(string[] args) { Console.WriteLine(LPC("aaaaabaababbaabaaababababababababbbbaaaaa")); }}// This code is contributed by prince |
Output
13
Time Complexity: O(N^3), where N is the length of the input string.
Auxiliary Space: O(N^2)
This article is contributed by Aditya Nihal Kumar Singh. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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