Palindrome pair in an array of words (or strings)
Last Updated :
23 Jul, 2025
Given an array of strings arr[] of size n, the task is to find if there exists two strings arr[i] and arr[j] such that arr[i]+arr[j] is a palindrome i.e the concatenation of string arr[i] and arr[j] results into a palindrome.
Examples:
Input: arr[] = ["geekf", "geeks", "or", "keeg", "abc", "bc"]
Output: True
Explanation: Strings "geekf" and "keeg" can be concatenated to form a palindromic string "geekfkeeg" .
Input: arr[] = ["abc", "xyxcba", "geekst", "or", "keeg", "bc"]
Output: True
Explanation: Strings "abc" and "xyxcba" can be concatenated to form a palindromic string "abcxyxcba" .
Input: arr[] = ["abc", "ab", "xyz"]
Output: False
Explanation: No palindromic string can be formed by concatenating any two of the given strings.
[Naive Approach] Using Nested Loop
The idea is to iteratively generate all possible pairs of strings using the nested loops and check if any of them is a palindrome.
C++
// C++ program to check if the given array
// contains a plaindromic pair of strings
#include<bits/stdc++.h>
using namespace std;
// Function to check if a string is palindrome
bool isPalindrome(string &s) {
int n = s.length();
// compare each character from starting
// with its corresponding character from last
for (int i = 0; i < n/2; i++ )
if (s[i] != s[n-i-1])
return false;
return true;
}
// Function to check if a palindrome pair exists
bool palindromePair(vector<string> &arr) {
// Consider each pair one by one
for (int i = 0; i < arr.size(); i++) {
for (int j = i + 1; j< arr.size(); j++) {
// concatenate both strings
string str = arr[i] + arr[j];
// check if str is palindrome
if (isPalindrome(str))
return true;
// check for other combination
str = arr[j] + arr[i];
if (isPalindrome(str))
return true;
}
}
return false;
}
int main() {
vector <string> arr = {"geekf", "geeks", "or", "keeg", "abc", "bc"};
if(palindromePair(arr))
cout << "True";
else
cout << "False";
return 0;
}
// Java program to check if the given array
// contains a palindromic pair of strings
import java.util.*;
class GfG {
// Function to check if a string is palindrome
static boolean isPalindrome(String s) {
int n = s.length();
// compare each character from starting
// with its corresponding character from last
for (int i = 0; i < n / 2; i++)
if (s.charAt(i) != s.charAt(n - i - 1))
return false;
return true;
}
// Function to check if a palindrome pair exists
static boolean palindromePair(String[] arr) {
// Consider each pair one by one
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < arr.length; j++) {
// concatenate both strings
String str = arr[i] + arr[j];
// check if str is palindrome
if (isPalindrome(str))
return true;
// check for other combination
str = arr[j] + arr[i];
if (isPalindrome(str))
return true;
}
}
return false;
}
public static void main(String[] args) {
String[] arr = {"geekf", "geeks", "or", "keeg", "abc", "bc"};
if (palindromePair(arr))
System.out.println("True");
else
System.out.println("False");
}
}
# Python program to check if the given array
# contains a palindromic pair of strings
# Function to check if a string is palindrome
def isPalindrome(s):
n = len(s)
# compare each character from starting
# with its corresponding character from last
for i in range(n // 2):
if s[i] != s[n - i - 1]:
return False
return True
# Function to check if a palindrome pair exists
def palindromePair(arr):
# Consider each pair one by one
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
# concatenate both strings
strVal = arr[i] + arr[j]
# check if str_val is palindrome
if isPalindrome(strVal):
return True
# check for other combination
strVal = arr[j] + arr[i]
if isPalindrome(strVal):
return True
return False
if __name__ == "__main__":
arr = ["geekf", "geeks", "or", "keeg", "abc", "bc"]
if palindromePair(arr):
print("True")
else:
print("False")
// C# program to check if the given array
// contains a palindromic pair of strings
using System;
class GfG {
// Function to check if a string is palindrome
static bool isPalindrome(string s) {
int n = s.Length;
// compare each character from starting
// with its corresponding character from last
for (int i = 0; i < n / 2; i++)
if (s[i] != s[n - i - 1])
return false;
return true;
}
// Function to check if a palindrome pair exists
static bool palindromePair(string[] arr) {
// Consider each pair one by one
for (int i = 0; i < arr.Length; i++) {
for (int j = i + 1; j < arr.Length; j++) {
// concatenate both strings
string str = arr[i] + arr[j];
// check if str is palindrome
if (isPalindrome(str))
return true;
// check for other combination
str = arr[j] + arr[i];
if (isPalindrome(str))
return true;
}
}
return false;
}
static void Main() {
string[] arr = { "geekf", "geeks", "or", "keeg", "abc", "bc" };
if (palindromePair(arr))
Console.WriteLine("True");
else
Console.WriteLine("False");
}
}
// JavaScript program to check if the given array
// contains a palindromic pair of strings
// Function to check if a string is palindrome
function isPalindrome(s) {
let n = s.length;
// compare each character from starting
// with its corresponding character from last
for (let i = 0; i < Math.floor(n / 2); i++)
if (s[i] !== s[n - i - 1])
return false;
return true;
}
// Function to check if a palindrome pair exists
function palindromePair(arr) {
// Consider each pair one by one
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
// concatenate both strings
let str = arr[i] + arr[j];
// check if str is palindrome
if (isPalindrome(str))
return true;
// check for other combination
str = arr[j] + arr[i];
if (isPalindrome(str))
return true;
}
}
return false;
}
// Driver code
let arr = ["geekf", "geeks", "or", "keeg", "abc", "bc"];
if (palindromePair(arr))
console.log("True");
else
console.log("False");
Time Complexity: O(n2 * k), here n is the number of words in the array arr[] and k is the length of the longest concatenated string.
Auxiliary Space: O(1)
[Expected Approach] Using Trie Data Structure
The idea is to use Trie Data Structure to store all the strings and efficiently search for palindromic pairs. To do so, initialize an empty Trie and store the strings of arr[] in reverse order. Also for each index ind of string arr[i], store if substring arr[i][0] to arr[i][ind] is a palindrome.
Now, traverse the array arr[] again, and for each string arr[i], check if it is present in Trie, if so, return true. Otherwise, if the string arr[i] exists partially in Trie, check if remaining part is palindrome or not, if so, return true, else move to the next string. If no pair of string is found, return false.
C++
// C++ program to check if the given array
// contains a plaindromic pair of strings
#include<bits/stdc++.h>
using namespace std;
// Trie node
class TrieNode {
public:
// pointer array for 26 letters
vector<TrieNode*> children;
//indices of palindromic words
vector<int> indices;
// index of word in array
int idx;
TrieNode() : idx(-1) {
// initialize children array to NULL
children.resize(26, nullptr);
}
};
// Function to check if a string is palindrome
bool isPalindrome(string str, int i, int j) {
// compare each character from starting
// with its corresponding character from last
while (i < j) {
if (str[i] != str[j])
return false;
i++, j--;
}
return true;
}
// If not present, inserts reverse of word into Trie. If
// the word is prefix of a Trie node, just mark leaf node
void insert(TrieNode* root, string word, int idx) {
TrieNode *node = root;
// Start traversing word from the last
for (int i = word.length() - 1; i >= 0; i--) {
// If current word is palindrome till this
// i, store index of current word.
if (isPalindrome(word, 0, i))
(node->indices).push_back(idx);
// If it is not available in Trie
// then store it
int index = word[i] - 'a';
if (!node->children[index])
node->children[index] = new TrieNode();
// move to new TrieNode
node = node->children[index];
}
node->idx = idx;
node->indices.push_back(idx);
}
// Returns true if word presents in Trie, else false
bool search(TrieNode *root, string word, int idx) {
TrieNode *node = root;
for (int i = 0; i < word.length() && node; i++) {
int index = word[i] - 'a';
// If it is present also check upto
// which index it is palindrome
if (node->idx >= 0 && node->idx != idx &&
isPalindrome(word, i, word.size()-1))
return true;
node = node->children[index];
}
if(node) {
for (int i : node->indices) {
if (idx != i)
return true;
}
}
return false;
}
// Function to check if a palindrome pair exists
bool palindromePair(vector<string> &arr) {
// Construct trie
TrieNode *root = new TrieNode();
for (int i = 0; i < arr.size(); i++)
insert(root, arr[i], i);
// Search for different keys
for (int i = 0; i < arr.size(); i++) {
if (search(root, arr[i], i))
return true;
}
return false;
}
int main() {
vector <string> arr = {"geekf", "geeks", "or", "keeg", "abc", "bc"};
if(palindromePair(arr))
cout << "True";
else
cout << "False";
return 0;
}
// Java program to check if the given array
// contains a palindromic pair of strings
import java.util.*;
class TrieNode {
// pointer array for 26 letters
TrieNode[] children;
// indices of palindromic words
List<Integer> indices;
// index of word in array
int idx;
TrieNode() {
this.idx = -1;
// initialize children array to NULL
this.children = new TrieNode[26];
this.indices = new ArrayList<>();
}
}
class GFG {
// Function to check if a string is palindrome
static boolean isPalindrome(String str, int i, int j) {
// compare each character from starting
// with its corresponding character from last
while (i < j) {
if (str.charAt(i) != str.charAt(j))
return false;
i++; j--;
}
return true;
}
// If not present, inserts reverse of word into Trie.
// If the word is prefix of a Trie node, just mark leaf node
static void insert(TrieNode root, String word, int idx) {
TrieNode node = root;
// Start traversing word from the last
for (int i = word.length() - 1; i >= 0; i--) {
// If current word is palindrome till this
// i, store index of current word.
if (isPalindrome(word, 0, i))
node.indices.add(idx);
// If it is not available in Trie
// then store it
int index = word.charAt(i) - 'a';
if (node.children[index] == null)
node.children[index] = new TrieNode();
// move to new TrieNode
node = node.children[index];
}
node.idx = idx;
node.indices.add(idx);
}
// Returns true if word presents in Trie, else false
static boolean search(TrieNode root, String word, int idx) {
TrieNode node = root;
for (int i = 0; i < word.length() && node != null; i++) {
int index = word.charAt(i) - 'a';
// If it is present also check upto
// which index it is palindrome
if (node.idx >= 0 && node.idx != idx &&
isPalindrome(word, i, word.length()-1))
return true;
node = node.children[index];
}
if (node != null) {
for (int i : node.indices) {
if (idx != i)
return true;
}
}
return false;
}
// Function to check if a palindrome pair exists
static boolean palindromePair(String[] arr) {
// Construct trie
TrieNode root = new TrieNode();
for (int i = 0; i < arr.length; i++)
insert(root, arr[i], i);
// Search for different keys
for (int i = 0; i < arr.length; i++) {
if (search(root, arr[i], i))
return true;
}
return false;
}
public static void main(String[] args) {
String[] arr = {"geekf", "geeks", "or", "keeg", "abc", "bc"};
if (palindromePair(arr))
System.out.println("True");
else
System.out.println("False");
}
}
# Python program to check if the given array
# contains a palindromic pair of strings
# Trie node
class TrieNode:
def __init__(self):
# pointer array for 26 letters
self.children = [None] * 26
# indices of palindromic words
self.indices = []
# index of word in array
self.idx = -1
# Function to check if a string is palindrome
def isPalindrome(word, i, j):
# compare each character from starting
# with its corresponding character from last
while i < j:
if word[i] != word[j]:
return False
i += 1
j -= 1
return True
# If not present, inserts reverse of word into Trie.
# If the word is prefix of a Trie node, just mark leaf node
def insert(root, word, idx):
node = root
# Start traversing word from the last
for i in range(len(word) - 1, -1, -1):
# If current word is palindrome till this
# i, store index of current word.
if isPalindrome(word, 0, i):
node.indices.append(idx)
index = ord(word[i]) - ord('a')
if not node.children[index]:
node.children[index] = TrieNode()
node = node.children[index]
node.idx = idx
node.indices.append(idx)
# Returns true if word presents in Trie, else false
def search(root, word, idx):
node = root
for i in range(len(word)):
index = ord(word[i]) - ord('a')
# If it is present also check up to
# which index it is palindrome
if node.idx >= 0 and node.idx != idx and \
isPalindrome(word, i, len(word) - 1):
return True
if not node.children[index]:
return False
node = node.children[index]
return any(i != idx for i in node.indices)
# Function to check if a palindrome pair exists
def palindromePair(arr):
root = TrieNode()
# Construct trie
for i in range(len(arr)):
insert(root, arr[i], i)
# Search for different keys
for i in range(len(arr)):
if search(root, arr[i], i):
return True
return False
if __name__ == "__main__":
arr = ["geekf", "geeks", "or", "keeg", "abc", "bc"]
if palindromePair(arr):
print("True")
else:
print("False")
// C# program to check if the given array
// contains a palindromic pair of strings
using System;
using System.Collections.Generic;
class TrieNode {
// pointer array for 26 letters
public TrieNode[] children;
// indices of palindromic words
public List<int> indices;
// index of word in array
public int idx;
public TrieNode() {
// initialize children array to null
children = new TrieNode[26];
indices = new List<int>();
idx = -1;
}
}
class GFG {
// Function to check if a string is palindrome
static bool isPalindrome(string word, int i, int j) {
// compare each character from starting
// with its corresponding character from last
while (i < j) {
if (word[i] != word[j])
return false;
i++; j--;
}
return true;
}
// If not present, inserts reverse of word into Trie.
// If the word is prefix of a Trie node, just mark leaf node
static void insert(TrieNode root, string word, int idx) {
TrieNode node = root;
// Start traversing word from the last
for (int i = word.Length - 1; i >= 0; i--) {
// If current word is palindrome till this
// i, store index of current word.
if (isPalindrome(word, 0, i))
node.indices.Add(idx);
// If it is not available in Trie
// then store it
int index = word[i] - 'a';
if (node.children[index] == null)
node.children[index] = new TrieNode();
// move to new TrieNode
node = node.children[index];
}
node.idx = idx;
node.indices.Add(idx);
}
// Returns true if word presents in Trie, else false
static bool search(TrieNode root, string word, int idx) {
TrieNode node = root;
for (int i = 0; i < word.Length && node != null; i++) {
int index = word[i] - 'a';
// If it is present also check up to
// which index it is palindrome
if (node.idx >= 0 && node.idx != idx &&
isPalindrome(word, i, word.Length - 1))
return true;
node = node.children[index];
}
if (node != null) {
foreach (int i in node.indices) {
if (idx != i)
return true;
}
}
return false;
}
// Function to check if a palindrome pair exists
static bool palindromePair(string[] arr) {
// Construct trie
TrieNode root = new TrieNode();
for (int i = 0; i < arr.Length; i++)
insert(root, arr[i], i);
// Search for different keys
for (int i = 0; i < arr.Length; i++) {
if (search(root, arr[i], i))
return true;
}
return false;
}
static void Main() {
string[] arr = { "geekf", "geeks", "or", "keeg", "abc", "bc" };
if (palindromePair(arr))
Console.WriteLine("True");
else
Console.WriteLine("False");
}
}
// JavaScript program to check if the given array
// contains a palindromic pair of strings
class TrieNode {
constructor() {
// pointer array for 26 letters
this.children = new Array(26).fill(null);
// indices of palindromic words
this.indices = [];
// index of word in array
this.idx = -1;
}
}
// Function to check if a string is palindrome
function isPalindrome(word, i, j) {
// compare each character from starting
// with its corresponding character from last
while (i < j) {
if (word[i] !== word[j])
return false;
i++; j--;
}
return true;
}
// If not present, inserts reverse of word into Trie.
// If the word is prefix of a Trie node, just mark leaf node
function insert(root, word, idx) {
let node = root;
// Start traversing word from the last
for (let i = word.length - 1; i >= 0; i--) {
// If current word is palindrome till this
// i, store index of current word.
if (isPalindrome(word, 0, i))
node.indices.push(idx);
// If it is not available in Trie
// then store it
let index = word.charCodeAt(i) - 'a'.charCodeAt(0);
if (!node.children[index])
node.children[index] = new TrieNode();
// move to new TrieNode
node = node.children[index];
}
node.idx = idx;
node.indices.push(idx);
}
// Returns true if word presents in Trie, else false
function search(root, word, idx) {
let node = root;
for (let i = 0; i < word.length && node; i++) {
let index = word.charCodeAt(i) - 'a'.charCodeAt(0);
// If it is present also check upto
// which index it is palindrome
if (node.idx >= 0 && node.idx !== idx &&
isPalindrome(word, i, word.length - 1))
return true;
node = node.children[index];
}
if (node) {
for (let i of node.indices) {
if (idx !== i)
return true;
}
}
return false;
}
// Function to check if a palindrome pair exists
function palindromePair(arr) {
// Construct trie
let root = new TrieNode();
for (let i = 0; i < arr.length; i++)
insert(root, arr[i], i);
// Search for different keys
for (let i = 0; i < arr.length; i++) {
if (search(root, arr[i], i))
return true;
}
return false;
}
// Driver code
let arr = ["geekf", "geeks", "or", "keeg", "abc", "bc"];
if (palindromePair(arr))
console.log("True");
else
console.log("False");
Time Complexity: O(n * k2), here n is the number of strings in arr[] and k is the length of the longest string. For each string, we are checking if the substring arr[i][0..ind] is a palindrome (0 <= ind <= k), that takes (n * k * k) time.
Auxiliary Space: O(n * k), to store all the strings in the Trie.
[Alternate Approach] Using HashMap
The above approach can also be implemented using HashMap instead of Trie.
C++
// C++ program to check if the given array
// contains a plaindromic pair of strings
#include <bits/stdc++.h>
using namespace std;
// Function to check if a string is palindrome
bool isPalindrome(string &s) {
int n = s.length();
// compare each character from starting
// with its corresponding character from last
for (int i = 0; i < n/2; i++ )
if (s[i] != s[n-i-1])
return false;
return true;
}
// Function to check if a palindrome pair exists
bool palindromePair(vector<string> &arr) {
// to store all substrings of all words
// along with indices in reverse order
unordered_map<string, int> str;
// to store all reverse of all words
for (int i = 0; i < arr.size(); i++) {
string word = arr[i];
reverse(word.begin(), word.end());
str[word] = i;
}
// enumerating over all words and
// for each character of them
for (int i = 0; i < arr.size(); i++) {
// to store the left substring of arr[i]
string left = "";
for (int j = 0; j < arr[i].size(); j++) {
// add currrent character to word
left += arr[i][j];
// extract right substring
string right = arr[i].substr(j+1);
// if word is not empty and is palindrome
// and right is present in the map
if(!left.empty() && isPalindrome(left) && str.count(right) && str[right] != i)
return true;
// check the same by swapping word and right
if(isPalindrome(right) && str.count(left) && str[left] != i)
return true;
}
}
return false;
}
int main() {
vector<string> arr = {"geekf", "geeks", "or", "keeg", "abc", "bc"};
if (palindromePair(arr))
cout << "True";
else
cout << "False";
return 0;
}
// Java program to check if the given array
// contains a palindromic pair of strings
import java.util.*;
class GFG {
// Function to check if a string is palindrome
static boolean isPalindrome(String s) {
int n = s.length();
// compare each character from starting
// with its corresponding character from last
for (int i = 0; i < n / 2; i++)
if (s.charAt(i) != s.charAt(n - i - 1))
return false;
return true;
}
// Function to check if a palindrome pair exists
static boolean palindromePair(String[] arr) {
// to store all substrings of all words
// along with indices in reverse order
Map<String, Integer> str = new HashMap<>();
// to store all reverse of all words
for (int i = 0; i < arr.length; i++) {
String word = new StringBuilder(arr[i]).reverse().toString();
str.put(word, i);
}
// enumerating over all words and
// for each character of them
for (int i = 0; i < arr.length; i++) {
// to store the left substring of arr[i]
String left = "";
for (int j = 0; j < arr[i].length(); j++) {
// add current character to word
left += arr[i].charAt(j);
// extract right substring
String right = arr[i].substring(j + 1);
// if word is not empty and is palindrome
// and right is present in the map
if (!left.isEmpty() && isPalindrome(left)
&& str.containsKey(right) && str.get(right) != i)
return true;
// check the same by swapping word and right
if (isPalindrome(right) && str.containsKey(left)
&& str.get(left) != i)
return true;
}
}
return false;
}
public static void main(String[] args) {
String[] arr = {"geekf", "geeks", "or", "keeg", "abc", "bc"};
if (palindromePair(arr))
System.out.println("True");
else
System.out.println("False");
}
}
# Python program to check if the given array
# contains a palindromic pair of strings
# Function to check if a string is palindrome
def isPalindrome(s):
n = len(s)
# compare each character from starting
# with its corresponding character from last
for i in range(n // 2):
if s[i] != s[n - i - 1]:
return False
return True
# Function to check if a palindrome pair exists
def palindromePair(arr):
# to store all substrings of all words
# along with indices in reverse order
strMap = {}
# to store all reverse of all words
for i in range(len(arr)):
word = arr[i][::-1]
strMap[word] = i
# enumerating over all words and
# for each character of them
for i in range(len(arr)):
# to store the left substring of arr[i]
left = ""
for j in range(len(arr[i])):
# add current character to word
left += arr[i][j]
# extract right substring
right = arr[i][j + 1:]
# if word is not empty and is palindrome
# and right is present in the map
if left and isPalindrome(left) and \
right in strMap and strMap[right] != i:
return True
# check the same by swapping word and right
if isPalindrome(right) and \
left in strMap and strMap[left] != i:
return True
return False
if __name__ == "__main__":
arr = ["geekf", "geeks", "or", "keeg", "abc", "bc"]
if palindromePair(arr):
print("True")
else:
print("False")
// C# program to check if the given array
// contains a palindromic pair of strings
using System;
using System.Collections.Generic;
class GFG {
// Function to check if a string is palindrome
static bool isPalindrome(string s) {
int n = s.Length;
// compare each character from starting
// with its corresponding character from last
for (int i = 0; i < n / 2; i++)
if (s[i] != s[n - i - 1])
return false;
return true;
}
// Function to check if a palindrome pair exists
static bool palindromePair(string[] arr) {
// to store all substrings of all words
// along with indices in reverse order
Dictionary<string, int> str = new Dictionary<string, int>();
// to store all reverse of all words
for (int i = 0; i < arr.Length; i++) {
char[] wordArr = arr[i].ToCharArray();
Array.Reverse(wordArr);
string word = new string(wordArr);
str[word] = i;
}
// enumerating over all words and
// for each character of them
for (int i = 0; i < arr.Length; i++) {
// to store the left substring of arr[i]
string left = "";
for (int j = 0; j < arr[i].Length; j++) {
// add current character to word
left += arr[i][j];
// extract right substring
string right = arr[i].Substring(j + 1);
// if word is not empty and is palindrome
// and right is present in the map
if (!string.IsNullOrEmpty(left) && isPalindrome(left)
&& str.ContainsKey(right) && str[right] != i)
return true;
// check the same by swapping word and right
if (isPalindrome(right) && str.ContainsKey(left)
&& str[left] != i)
return true;
}
}
return false;
}
public static void Main() {
string[] arr = { "geekf", "geeks", "or", "keeg", "abc", "bc" };
if (palindromePair(arr))
Console.WriteLine("True");
else
Console.WriteLine("False");
}
}
// JavaScript program to check if the given array
// contains a palindromic pair of strings
// Function to check if a string is palindrome
function isPalindrome(s) {
let n = s.length;
// compare each character from starting
// with its corresponding character from last
for (let i = 0; i < Math.floor(n / 2); i++)
if (s[i] !== s[n - i - 1])
return false;
return true;
}
// Function to check if a palindrome pair exists
function palindromePair(arr) {
// to store all substrings of all words
// along with indices in reverse order
let str = new Map();
// to store all reverse of all words
for (let i = 0; i < arr.length; i++) {
let word = arr[i].split("").reverse().join("");
str.set(word, i);
}
// enumerating over all words and
// for each character of them
for (let i = 0; i < arr.length; i++) {
// to store the left substring of arr[i]
let left = "";
for (let j = 0; j < arr[i].length; j++) {
// add current character to word
left += arr[i][j];
// extract right substring
let right = arr[i].substring(j + 1);
// check palindrome conditions
if (left && isPalindrome(left) && str.has(right)
&& str.get(right) !== i)
return true;
if (isPalindrome(right) && str.has(left)
&& str.get(left) !== i)
return true;
}
}
return false;
}
// Driver code
let arr = ["geekf", "geeks", "or", "keeg", "abc", "bc"];
console.log(palindromePair(arr) ? "True" : "False");
Time Complexity: O(n * k2), here n is the number of strings in arr[] and k is the length of the longest string. For each string, we are checking if the substring arr[i][0..ind] is a palindrome (0 <= ind <= k), that takes (n * k * k) time.
Auxiliary Space: O(n * k), to store all the strings in the HashMap.
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