# Palindrome pair in an array of words (or strings)

Given a list of words, find if any of the two words can be joined to form a palindrome.

Examples:

```Input  : list[] = {"geekf", "geeks", "or",
"keeg", "abc", "bc"}
Output : Yes
There is a pair "geekf" and "keeg"

Input : list[] =  {"abc", "xyxcba", "geekst", "or",
"keeg", "bc"}
Output : Yes
There is a pair "abc" and "xyxcba"
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple approach:

```1- Consider each pair one by one.
2- Check if any of the pairs forms a palindrome
after concatenating them.
3- Return true, if any such pair exists.
4- Else, return false.
```

## C++

```// C++ program to find if there is a pair that
// can form a palindrome.
#include<bits/stdc++.h>
using namespace std;

// Utility function to check if a string is a
// palindrome
bool isPalindrome(string str)
{
int len = str.length();

// compare each character from starting
// with its corresponding character from last
for (int i = 0; i < len/2; i++ )
if (str[i] != str[len-i-1])
return false;

return true;
}

// Function to check if a palindrome pair exists
bool checkPalindromePair(vector <string> vect)
{
// Consider each pair one by one
for (int i = 0; i< vect.size()-1; i++)
{
for (int j = i+1; j< vect.size() ; j++)
{
string check_str = "";

// concatenate both strings
check_str = check_str + vect[i] + vect[j];

// check if the concatenated string is
// palindrome
if (isPalindrome(check_str))
return true;
}
}
return false;
}

// Driver code
int main()
{
vector <string> vect = {"geekf", "geeks", "or",
"keeg", "abc", "bc"};

checkPalindromePair(vect)? cout << "Yes" :
cout << "No";
return 0;
}
```

## Java

```// Java program to find if there is a pair that
// can form a palindrome.
import java.util.Arrays;
import java.util.List;
public class Palin_pair1 {

// Utility function to check if a string is a
// palindrome
static boolean isPalindrome(String str)
{
int len = str.length();

// compare each character from starting
// with its corresponding character from last
for (int i = 0; i < len/2; i++ )
if (str.charAt(i) != str.charAt(len-i-1))
return false;

return true;
}

// Function to check if a palindrome pair exists
static boolean checkPalindromePair(List<String> vect)
{
// Consider each pair one by one
for (int i = 0; i< vect.size()-1; i++)
{
for (int j = i+1; j< vect.size() ; j++)
{
String check_str = "";

// concatenate both strings
check_str = check_str + vect.get(i) + vect.get(j);

// check if the concatenated string is
// palindrome
if (isPalindrome(check_str))
return true;
}
}
return false;
}

// Driver code
public static void main(String args[])
{
List<String> vect = Arrays.asList("geekf", "geeks", "or",
"keeg", "abc", "bc");

if (checkPalindromePair(vect) == true)
System.out.println("Yes");
else
System.out.println("No");
}
}
//This code is contributed by Sumit Ghosh
```

Output:

``` Yes
```

Time Complexity : O(n2k)
Here n is the number of the words in the list and k is the maximum length that is checked for a palindrome.

Efficient method
It can be done in an efficient manner by using the Trie data structure. The idea is to maintain a Trie of the reverse of all words.

```1) Create an empty Trie.
2) Do following for every word:-
a) Insert reverse of current word.
b) Also store up to which index it is
a palindrome.
3) Traverse list of words again and do following
for every word.
a) If it is available in Trie then return true
b) If it is partially available
Check the remaining word is palindrome or not
If yes then return true that means a pair
forms a palindrome.
Note: Position upto which the word is palindrome
is stored because of these type of cases.
```

## C++

```// C++ program to check if there is a pair that
// of above method using Trie
#include<bits/stdc++.h>
using namespace std;
#define ARRAY_SIZE(a) sizeof(a)/sizeof(a[0])

// Alphabet size (# of symbols)
#define ALPHABET_SIZE (26)

// Converts key current character into index
// use only 'a' through 'z' and lower case
#define CHAR_TO_INDEX(c) ((int)c - (int)'a')

// Trie node
struct TrieNode
{
struct TrieNode *children[ALPHABET_SIZE];
vector<int> pos; // To store palindromic
// positions in str
int id;

// isLeaf is true if the node represents
// end of a word
bool isLeaf;
};

// Returns new Trie node (initialized to NULLs)
struct TrieNode *getNode(void)
{
struct TrieNode *pNode = new TrieNode;
pNode->isLeaf = false;
for (int i = 0; i < ALPHABET_SIZE; i++)
pNode->children[i] = NULL;

return pNode;
}

// Utility function to check if a string is a
// palindrome
bool isPalindrome(string str, int i, int len)
{
// compare each character from starting
// with its corresponding character from last
while (i < len)
{
if (str[i] != str[len])
return false;
i++, len--;
}

return true;
}

// If not present, inserts reverse of key into Trie. If
// the key is prefix of a Trie node, just mark leaf node
void insert(struct TrieNode* root, string key, int id)
{
struct TrieNode *pCrawl = root;

// Start traversing word from the last
for (int level = key.length()-1; level >=0; level--)
{
// If it is not available in Trie, then
// store it
int index = CHAR_TO_INDEX(key[level]);
if (!pCrawl->children[index])
pCrawl->children[index] = getNode();

// If current word is palindrome till this
// level, store index of current word.
if (isPalindrome(key, 0, level))
(pCrawl->pos).push_back(id);

pCrawl = pCrawl->children[index];
}
pCrawl->id = id;
pCrawl->pos.push_back(id);

// mark last node as leaf
pCrawl->isLeaf = true;
}

// Returns true if key presents in Trie, else false
void search(struct TrieNode *root, string key,
int id, vector<vector<int> > &result)
{
struct TrieNode *pCrawl = root;
for (int level = 0; level < key.length(); level++)
{
int index = CHAR_TO_INDEX(key[level]);

// If it is present also check upto which index
// it is palindrome
if (pCrawl->id >= 0 && pCrawl->id != id &&
isPalindrome(key, level, key.size()-1))
result.push_back({id, pCrawl->id});

// If not present then return
if (!pCrawl->children[index])
return;

pCrawl = pCrawl->children[index];
}

for (int i: pCrawl->pos)
{
if (i == id)
continue;
result.push_back({id, i});
}
}

// Function to check if a palindrome pair exists
bool checkPalindromePair(vector <string> vect)
{
// Construct trie
struct TrieNode *root = getNode();
for (int i = 0; i < vect.size(); i++)
insert(root, vect[i], i);

// Search for different keys
vector<vector<int> > result;
for (int i=0; i<vect.size(); i++)
{
search(root, vect[i], i, result);
if (result.size() > 0)
return true;
}

return false;
}

// Driver code
int main()
{
vector <string> vect = {"geekf", "geeks", "or",
"keeg", "abc", "bc"};

checkPalindromePair(vect)? cout << "Yes" :
cout << "No";
return 0;
}
```

## Java

```
//Java program to check if there is a pair that
//of above method using Trie
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Palin_pair2 {

// Alphabet size (# of symbols)
static final int ALPHABET_SIZE = 26;

// Trie node
static class TrieNode {
TrieNode[] children = new TrieNode[ALPHABET_SIZE];
List<Integer> pos; // To store palindromic
// positions in str
int id;

// isLeaf is true if the node represents
// end of a word
boolean isLeaf;

// constructor
public TrieNode() {
isLeaf = false;
pos = new ArrayList<>();
for (int i = 0; i < ALPHABET_SIZE; i++)
children[i] = null;
}
}

// Utility function to check if a string is a
// palindrome
static boolean isPalindrome(String str, int i, int len) {
// compare each character from starting
// with its corresponding character from last
while (i < len) {
if (str.charAt(i) != str.charAt(len))
return false;

i++;
len--;
}
return true;
}

// If not present, inserts reverse of key into Trie. If
// the key is prefix of a Trie node, just mark leaf node
static void insert(TrieNode root, String key, int id) {
TrieNode pCrawl = root;

// Start traversing word from the last
for (int level = key.length() - 1; level >= 0; level--) {
// If it is not available in Trie, then
// store it
int index = key.charAt(level) - 'a';
if (pCrawl.children[index] == null)
pCrawl.children[index] = new TrieNode();

// If current word is palindrome till this
// level, store index of current word.
if (isPalindrome(key, 0, level))

pCrawl = pCrawl.children[index];
}
pCrawl.id = id;

// mark last node as leaf
pCrawl.isLeaf = true;
}

// list to store result
static List<List<Integer>> result;

// Returns true if key presents in Trie, else false
static void search(TrieNode root, String key, int id) {
TrieNode pCrawl = root;
for (int level = 0; level < key.length(); level++) {
int index = key.charAt(level) - 'a';

// If it is present also check upto which index
// it is palindrome
if (pCrawl.id >= 0 && pCrawl.id != id
&& isPalindrome(key, level, key.length() - 1)) {
List<Integer> l = new ArrayList<>();
}

// If not present then return
if (pCrawl.children[index] == null)
return;

pCrawl = pCrawl.children[index];
}

for (int i : pCrawl.pos) {
if (i == id)
continue;
List<Integer> l = new ArrayList<>();
}
}

// Function to check if a palindrome pair exists
static boolean checkPalindromePair(List<String> vect) {

// Construct trie
TrieNode root = new TrieNode();
for (int i = 0; i < vect.size(); i++)
insert(root, vect.get(i), i);

// Search for different keys
result = new ArrayList<>();
for (int i = 0; i < vect.size(); i++) {
search(root, vect.get(i), i);

if (result.size() > 0)
return true;
}

return false;
}

// Driver code
public static void main(String args[]) {
List<String> vect = Arrays.asList("geekf", "geeks",
"or", "keeg", "abc", "bc");

if (checkPalindromePair(vect) == true)
System.out.println("Yes");
else
System.out.println("No");
}
}
//This code is contributed by Sumit Ghosh
```

Output:
```Yes
```

Time Complexity: O(nk2)
Where n is the number of words in the list and k is the maximum length that is checked for palindrome.

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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