Print all possible palindromic string formed using any pair of given strings

Given an array of strings arr[] containing N words, the task is to print all possible palindromic string by combining any two strings from the given array.

Input: arr[] = [“geekf”, “geeks”, “or”, “keeg”, “abc”, “ba”]
Output: [“geekfkeeg”, “geekskeeg”, “abcba”]  
Below pairs forms the palindromic string on combining:
1. “geekf” + “keeg” = “geekfkeeg”
2. “geeks” + “keeg” = “geekskeeg”
3. “abc” + “ba” = “abcba”

Input: arr[] = [“dcb”, “yz”, “xy”, “efg”, “yx”] 
Output: [“xyyx”, “yxxy”]
1. “xy” + “yz” = “xyyz”
2. “yx” + “xy” = “yxxy”


Naive Approach: The naive approach is to iterate all possible pairs in the given array of strings and check whether concatenation of the string forms palindromic or not. If yes then print that pair and check for the next pairs.

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using Hashing. Below are the steps:

  1. Store all the words in a Map with words as keys and indices as values.
  2. For each word(say str) in the arr[] break the string into string s1 and s2 such that s1 + s2 = str.
  3. After the above step there arise two cases:
    • Case 1: If s1 is a palindromic string and the reverse of s2 is present in the hash-map then we get the required pair (reverse of s2  + current word).
    • Case 2: If s2 is a palindromic string and if reverse s1 is present in the hash-map then we get a pair (current word  +  reverse of s1).
  4. Repeat the above steps for all the strings in the array.

Below is the implementation of the above approach:






// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check whether the string
// word is palindrome or not
bool ispalin(string word)
    if (word.length() == 1
        || word.empty()) {
        return true;
    int l = 0;
    int r = word.length() - 1;
    // Iterate word
    while (l <= r) {
        if (word[l] != word[r]) {
            return false;
    return true;
// Function to find the palindromicPairs
palindromicPairs(vector<string>& words)
    vector<string> output;
    if (words.size() == 0
        || words.size() == 1) {
        return output;
    // Insert all the strings with
    // their indices in the hash map
    unordered_map<string, int> mp;
    for (int i = 0; i < words.size(); i++) {
        mp[words[i]] = i;
    // Iterate over all the words
    for (int i = 0; i < words.size(); i++) {
        // If the word is empty then
        // we simply pair it will all the
        // words which are palindrome
        // present in the array
        // except the word itself
        if (words[i].empty()) {
            for (auto x : mp) {
                if (x.second == i) {
                if (ispalin(x.first)) {
        // Create all possible substrings
        // s1 and s2
        for (int j = 0;
             j < words[i].length(); j++) {
            string s1 = words[i].substr(0, j + 1);
            string s2 = words[i].substr(j + 1);
            // Case 1
            // If s1 is palindrome and
            // reverse of s2 is
            // present in hashmap at
            // index other than i
            if (ispalin(s1)) {
                string temp = s2;
                if (mp.count(s2) == 1
                    && mp[s2] != i) {
                    string ans = s2 + words[i];
                s2 = temp;
            // Case 2
            // If s2 is palindrome and
            // reverse of s1 is
            // present in hashmap
            // at index other than i
            if (ispalin(s2)) {
                string temp = s1;
                if (mp.count(s1) == 1
                    && mp[s1] != i) {
                    string ans = words[i] + s1;
                s1 = temp;
    // Return output
    return output;
// Driver Code
int main()
    vector<string> words;
    // Given array of words
    words = { "geekf", "geeks", "or",
              "keeg", "abc", "ba" };
    // Function call
    vector<string> result
        = palindromicPairs(words);
    // Print the palindromic strings
    // after combining them
    for (auto x : result) {
        cout << x << endl;
    return 0;




Time Complexity: O(N)
Auxiliary Space: O(N)

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