Longest Palindromic Substring using Palindromic Tree | Set 3
Given a string, find the longest substring which is a palindrome. For example, if the given string is “forBeginnerskeegfor”, the output should be “Beginnerskeeg”.
Prerequisite : Palindromic Tree | Longest Palindromic Substring
Structure of Palindromic Tree :
The palindromic Tree’s actual structure is close to the directed graph. It is actually a merged structure of two Trees that share some common nodes(see the figure below for a better understanding). Tree nodes store palindromic substrings of a given string by storing their indices. This tree consists of two types of edges:
- Insertion edge (weighted edge)
- Maximum Palindromic Suffix (un-weighted)
Insertion Edge :
Insertion edge from a node u to v with some weight x means that the node v is formed by inserting x at the front and end of the string at u. As ‘u’ is already a palindrome, hence the resulting string at node v will also be a palindrome. x will be a single character for every edge. Therefore, a node can have a max of 26 insertion edges (considering lower letter string).
Maximum Palindromic Suffix Edge :
As the name itself indicates that for a node this edge will point to its Maximum Palindromic Suffix String node. We will not be considering the complete string itself as the Maximum Palindromic Suffix as this will make no sense(self-loops). For simplicity purposes, we will call it Suffix edge(by which we mean maximum suffix except for the complete string). It is quite obvious that every node will have only 1 Suffix Edge as we will not store duplicate strings in the tree. We will create all the palindromic substrings and then return the last one we got since that would be the longest palindromic substring so far. Since Palindromic Tree stores the palindromes in order of arrival of a certain character, the Longest will always be at the last index of our tree array
Below is the implementation of the above approach :
C++
// C++ program to demonstrate working of // palindromic tree #include "bits/stdc++.h" using namespace std; #define MAXN 1000 struct Node { // store start and end indexes of current // Node inclusively int start, end; // stores length of substring int length; // stores insertion Node for all characters a-z int insertEdg[26]; // stores the Maximum Palindromic Suffix Node for // the current Node int suffixEdg; }; // two special dummy Nodes as explained above Node root1, root2; // stores Node information for constant time access Node tree[MAXN]; // Keeps track the current Node while insertion int currNode; string s; int ptr; void insert( int idx) { // STEP 1// /* search for Node X such that s[idx] X S[idx] is maximum palindrome ending at position idx iterate down the suffix link of currNode to find X */ int tmp = currNode; while ( true ) { int curLength = tree[tmp].length; if (idx - curLength >= 1 and s[idx] == s[idx - curLength - 1]) break ; tmp = tree[tmp].suffixEdg; } /* Now we have found X .... * X = string at Node tmp * Check : if s[idx] X s[idx] already exists or not*/ if (tree[tmp].insertEdg[s[idx] - 'a' ] != 0) { // s[idx] X s[idx] already exists in the tree currNode = tree[tmp].insertEdg[s[idx] - 'a' ]; return ; } // creating new Node ptr++; // making new Node as child of X with // weight as s[idx] tree[tmp].insertEdg[s[idx] - 'a' ] = ptr; // calculating length of new Node tree[ptr].length = tree[tmp].length + 2; // updating end point for new Node tree[ptr].end = idx; // updating the start for new Node tree[ptr].start = idx - tree[ptr].length + 1; // STEP 2// /* Setting the suffix edge for the newly created Node tree[ptr]. Finding some String Y such that s[idx] + Y + s[idx] is longest possible palindromic suffix for newly created Node.*/ tmp = tree[tmp].suffixEdg; // making new Node as current Node currNode = ptr; if (tree[currNode].length == 1) { // if new palindrome's length is 1 // making its suffix link to be null string tree[currNode].suffixEdg = 2; return ; } while ( true ) { int curLength = tree[tmp].length; if (idx - curLength >= 1 and s[idx] == s[idx - curLength - 1]) break ; tmp = tree[tmp].suffixEdg; } // Now we have found string Y // linking current Nodes suffix link with // s[idx]+Y+s[idx] tree[currNode].suffixEdg = tree[tmp].insertEdg[s[idx] - 'a' ]; } // driver program int main() { // initializing the tree root1.length = -1; root1.suffixEdg = 1; root2.length = 0; root2.suffixEdg = 1; tree[1] = root1; tree[2] = root2; ptr = 2; currNode = 1; // given string s = "forBeginnerskeegfor" ; int l = s.length(); for ( int i = 0; i < l; i++) insert(i); int maxx = 0; string result = "" ; for ( int i = 3; i <= ptr; i++) { string res = "" ; for ( int j = tree[i].start; j <= tree[i].end; j++) res += s[j]; if (res.size() > maxx) { maxx = res.size(); result = res; } } cout << result; return 0; } |
Java
// Nikunj Sonigara // Define a class named Main class Main { // Define a static nested class named Node static class Node { int start, end; // Represents the start and end indices of a palindrome substring int length; // Length of the palindrome substring int [] insertEdge = new int [ 26 ]; // Array to store indices of child nodes for each character int suffixEdge; // Index of the node representing the largest proper suffix palindrome } // Create two root nodes and an array to store nodes static Node root1 = new Node(); static Node root2 = new Node(); static Node[] tree = new Node[ 1000 ]; static int ptr; // Pointer to the current node being processed static int currNode; // Index of the current node in the tree static String s; // Input string // Main method public static void main(String[] args) { // Initialize the root nodes root1.length = - 1 ; root1.suffixEdge = 1 ; root2.length = 0 ; root2.suffixEdge = 1 ; // Create nodes and initialize the tree array for ( int i = 0 ; i < 1000 ; i++) { tree[i] = new Node(); } tree[ 1 ] = root1; tree[ 2 ] = root2; ptr = 2 ; currNode = 1 ; // Input string s = "forBeginnerskeegfor" ; int l = s.length(); // Build the palindrome tree for ( int i = 0 ; i < l; i++) { insert(i); } // Find the longest palindrome substring in the tree int maxx = 0 ; String result = "" ; for ( int i = 3 ; i <= ptr; i++) { StringBuilder res = new StringBuilder(); for ( int j = tree[i].start; j <= tree[i].end; j++) { res.append(s.charAt(j)); } if (res.length() > maxx) { maxx = res.length(); result = res.toString(); } } // Print the result System.out.println(result); } // Method to insert a character into the palindrome tree static void insert( int idx) { int tmp = currNode; // Traverse the tree to find the largest proper suffix palindrome while ( true ) { int curLength = tree[tmp].length; if (idx - curLength >= 1 && s.charAt(idx) == s.charAt(idx - curLength - 1 )) { break ; } tmp = tree[tmp].suffixEdge; } // Check if the current character already has an edge in the tree if (tree[tmp].insertEdge[s.charAt(idx) - 'a' ] != 0 ) { currNode = tree[tmp].insertEdge[s.charAt(idx) - 'a' ]; return ; } // Create a new node for the current character ptr++; tree[tmp].insertEdge[s.charAt(idx) - 'a' ] = ptr; tree[ptr].length = tree[tmp].length + 2 ; tree[ptr].end = idx; tree[ptr].start = idx - tree[ptr].length + 1 ; // Update the current node and handle special case for length 1 palindrome tmp = tree[tmp].suffixEdge; currNode = ptr; if (tree[currNode].length == 1 ) { tree[currNode].suffixEdge = 2 ; return ; } // Traverse the tree again to find the largest proper suffix palindrome while ( true ) { int curLength = tree[tmp].length; if (idx - curLength >= 1 && s.charAt(idx) == s.charAt(idx - curLength - 1 )) { break ; } tmp = tree[tmp].suffixEdge; } // Update the suffix edge for the current node tree[currNode].suffixEdge = tree[tmp].insertEdge[s.charAt(idx) - 'a' ]; } } |
Python3
# Python program to demonstrate working of # palindromic tree MAXN = 1000 class Node: def __init__( self ): # store start and end indexes of current Node inclusively self .start = 0 self .end = 0 # stores length of substring self .length = 0 # stores insertion Node for all characters a-z self .insertEdg = [ 0 ] * 26 # stores the Maximum Palindromic Suffix Node for the current Node self .suffixEdg = 0 # two special dummy Nodes as explained above root1 = Node() root2 = Node() # stores Node information for constant time access tree = [Node() for i in range (MAXN)] # Keeps track the current Node while insertion currNode = 0 s = "" ptr = 0 def insert(idx): global currNode, ptr # STEP 1 # search for Node X such that s[idx] X S[idx] is maximum palindrome ending at position idx # iterate down the suffix link of currNode to find X tmp = currNode while True : curLength = tree[tmp].length if idx - curLength > = 1 and s[idx] = = s[idx - curLength - 1 ]: break tmp = tree[tmp].suffixEdg # Now we have found X.... # X = string at Node tmp # Check: if s[idx] X s[idx] already exists or not if tree[tmp].insertEdg[ ord (s[idx]) - ord ( 'a' )] ! = 0 : # s[idx] X s[idx] already exists in the tree currNode = tree[tmp].insertEdg[ ord (s[idx]) - ord ( 'a' )] return # creating new Node ptr + = 1 # making new Node as child of X with weight as s[idx] tree[tmp].insertEdg[ ord (s[idx]) - ord ( 'a' )] = ptr # calculating length of new Node tree[ptr].length = tree[tmp].length + 2 # updating end point for new Node tree[ptr].end = idx # updating the start for new Node tree[ptr].start = idx - tree[ptr].length + 1 # STEP 2 # Setting the suffix edge for the newly created Node tree[ptr]. Finding some String Y # such that s[idx] + Y + s[idx] is longest possible palindromic suffix for newly created Node. tmp = tree[tmp].suffixEdg # making new Node as current Node currNode = ptr if tree[currNode].length = = 1 : # if new palindrome's length is 1, making its suffix link to be null string tree[currNode].suffixEdg = 2 return while True : curLength = tree[tmp].length if idx - curLength > = 1 and s[idx] = = s[idx - curLength - 1 ]: break tmp = tree[tmp].suffixEdg # Now we have found string Y # linking current Nodes suffix link with s[idx]+Y+s[idx] tree[currNode].suffixEdg = tree[tmp].insertEdg[ ord (s[idx]) - ord ( 'a' )] # driver program # initializing the tree root1.length = - 1 root1.suffixEdg = 1 root2.length = 0 root2.suffixEdg = 1 tree[ 1 ] = root1 tree[ 2 ] = root2 ptr = 2 currNode = 1 # given string s = "forBeginnerskeegfor" l = len (s) for i in range (l): insert(i) maxx = 0 result = "" for i in range ( 3 , ptr + 1 ): res = "" for j in range (tree[i].start, tree[i].end + 1 ): res + = s[j] if len (res) > maxx: maxx = len (res) result = res print (result) |
C#
using System; class PalindromicTree { const int MAXN = 1000; // Node structure struct Node { // Store start and end indexes of current Node // inclusively public int start, end; // Store length of substring public int length; // Store insertion Node for all characters a-z public int [] insertEdge; // Store the Maximum Palindromic Suffix Node for the // current Node public int suffixEdge; } // Two special dummy Nodes as explained above static Node root1, root2; // Store Node information for constant time access static Node[] tree; // Keep track of the current Node while insertion static int currNode; static string s; static int ptr; // Function to insert a character into the Palindromic // Tree static void Insert( int idx) { // STEP 1 /* Search for Node X such that s[idx] X s[idx] * is the maximum palindrome ending at position idx. * Iterate down the suffix link of currNode to find * X. */ int tmp = currNode; while ( true ) { int curLength = tree[tmp].length; if (idx - curLength >= 1 && s[idx] == s[idx - curLength - 1]) break ; tmp = tree[tmp].suffixEdge; } /* Now we have found X... * X = string at Node tmp * Check: if s[idx] X s[idx] already exists or not */ if (tree[tmp].insertEdge[s[idx] - 'a' ] != 0) { // s[idx] X s[idx] already exists in the tree currNode = tree[tmp].insertEdge[s[idx] - 'a' ]; return ; } // Creating a new Node ptr++; // Initializing the insertEdge array for the new // Node tree[ptr].insertEdge = new int [26]; // Making the new Node as a child of X with weight // as s[idx] tree[tmp].insertEdge[s[idx] - 'a' ] = ptr; // Calculating the length of the new Node tree[ptr].length = tree[tmp].length + 2; // Updating the end point for the new Node tree[ptr].end = idx; // Updating the start for the new Node tree[ptr].start = idx - tree[ptr].length + 1; // STEP 2 /* Setting the suffix edge for the newly created * Node tree[ptr]. Finding some String Y such that * s[idx] + Y + s[idx] is the longest possible * palindromic suffix for the newly created Node. */ tmp = tree[tmp].suffixEdge; // Making the new Node as the current Node currNode = ptr; if (tree[currNode].length == 1) { // If the new palindrome's length is 1, // making its suffix link to be a null string tree[currNode].suffixEdge = 2; return ; } while ( true ) { int curLength = tree[tmp].length; if (idx - curLength >= 1 && s[idx] == s[idx - curLength - 1]) break ; tmp = tree[tmp].suffixEdge; } // Now we have found string Y // Linking the current Node's suffix link with // s[idx]+Y+s[idx] tree[currNode].suffixEdge = tree[tmp].insertEdge[s[idx] - 'a' ]; } // Driver program static void Main() { // Initializing the tree root1.length = -1; root1.suffixEdge = 1; root1.insertEdge = new int [26]; root2.length = 0; root2.suffixEdge = 1; root2.insertEdge = new int [26]; tree = new Node[MAXN]; tree[1] = root1; tree[2] = root2; ptr = 2; currNode = 1; // Given string s = "forBeginnerskeegfor" ; int l = s.Length; for ( int i = 0; i < l; i++) Insert(i); int maxx = 0; string result = "" ; for ( int i = 3; i <= ptr; i++) { string res = "" ; for ( int j = tree[i].start; j <= tree[i].end; j++) res += s[j]; if (res.Length > maxx) { maxx = res.Length; result = res; } } Console.WriteLine(result); } } |
Javascript
// JavaScript program to demonstrate working of // palindromic tree const MAXN = 1000; class Node { constructor() { this .start = 0; // store start and end indexes of current Node inclusively this .end = 0; this .length = 0; // stores length of substring this .insertEdg = Array(26).fill(0); // stores insertion Node for all characters a-z this .suffixEdg = 0; // stores the Maximum Palindromic Suffix Node for the current Node } } const root1 = new Node(); // two special dummy Nodes as explained above const root2 = new Node(); const tree = Array(MAXN).fill().map(() => new Node()); // stores Node information for constant time access let currNode = 0; let s = "" ; let ptr = 0; function insert(idx) { let tmp; // STEP 1 // search for Node X such that s[idx] X S[idx] is maximum palindrome ending at position idx // iterate down the suffix link of currNode to find X tmp = currNode; while ( true ) { const curLength = tree[tmp].length; if (idx - curLength >= 1 && s[idx] === s[idx - curLength - 1]) { break ; } tmp = tree[tmp].suffixEdg; } // Now we have found X.... // X = string at Node tmp // Check: if s[idx] X s[idx] already exists or not if (tree[tmp].insertEdg[s.charCodeAt(idx) - 97] !== 0) { // s[idx] X s[idx] already exists in the tree currNode = tree[tmp].insertEdg[s.charCodeAt(idx) - 97]; return ; } // creating new Node ptr += 1; // making new Node as child of X with weight as s[idx] tree[tmp].insertEdg[s.charCodeAt(idx) - 97] = ptr; // calculating length of new Node tree[ptr].length = tree[tmp].length + 2; // updating end point for new Node tree[ptr].end = idx; // updating the start for new Node tree[ptr].start = idx - tree[ptr].length + 1; // STEP 2 // Setting the suffix edge for the newly created Node tree[ptr]. Finding some String Y // such that s[idx] + Y + s[idx] is longest possible palindromic suffix for newly created Node. tmp = tree[tmp].suffixEdg; // making new Node as current Node currNode = ptr; if (tree[currNode].length === 1) { // if new palindrome's length is 1, making its suffix link to be null string tree[currNode].suffixEdg = 2; return ; } while ( true ) { const curLength = tree[tmp].length; if (idx - curLength >= 1 && s[idx] === s[idx - curLength - 1]) { break ; } tmp = tree[tmp].suffixEdg; } // Now we have found string Y // linking current Nodes suffix link with s[idx]+Y+s[idx] tree[currNode].suffixEdg = tree[tmp].insertEdg[s.charCodeAt(idx) - 97]; } // initializing the tree // const root1 = new Node(); root1.length = -1; root1.suffixEdg = 1; // const root2 = new Node(); root2.length = 0; root2.suffixEdg = 1; tree[1] = root1 tree[2] = root2 ptr = 2; currNode = 1; s = "forBeginnerskeegfor" ; const l = s.length; for (let i = 0; i < l; i++) { insert(i); } let maxx = 0; let result = "" ; for (let i = 3; i <= ptr; i++) { let res = "" ; for (let j = tree[i].start; j <= tree[i].end; j++) { res += s[j]; } if (res.length > maxx) { maxx = res.length; result = res; } } console.log(result); // Contributed by adityasha4x71 |
Beginnerskeeg
Time Complexity:
The time complexity for the building process will be O(k*n), here “n” is the length of the string and ‘k‘ is the extra iterations required to find the string X and string Y in the suffix links every time we insert a character.
Let’s try to approximate the constant ‘k’. We shall consider a worst case like s = “aaaaaabcccccccdeeeeeeeeef”. In this case for similar streak of continuous characters it will take extra 2 iterations per index to find both string X and Y in the suffix links , but as soon as it reaches some index i such that s[i]!=s[i-1] the left most pointer for the maximum length suffix will reach its rightmost limit. Therefore, for all i when s[i]!=s[i-1] , it will cost in total n iterations(summing over each iteration) and for rest i when s[i]==s[i-1] it takes 2 iteration which sums up over all such i and takes 2*n iterations. Hence, approximately our complexity in this case will be O(3*n) ~ O(n). So, we can roughly say that the constant factor ‘k’ will be very less. Therefore, we can consider the overall complexity to be linear O(length of string). You may refer the reference links for better understanding.
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