# Count pair of strings whose concatenation of substrings form a palindrome

Given an array of strings arr[], the task is to count the pair of strings whose concatenation of substrings form a palindrome.
Examples:

Input: arr[] = {“gfg”, “gfg”}
Output:
Explanation:
One possible way of choosing s1 and s2 is s1 = “gf”, s2 = “g” such that s1 + s2 i.e “gfg” is a palindrome.
Input: arr[] = {“abc”, B = “def”}
Output:

Approach: The key observation in the problem is if both strings have at least one common character let’s say ‘c’ then we can form a palindromic string. Therefore, check for all the pairs in the array that there is a common character in the string or not.
Below is the implementation of the above approach:

## C++

 `// C++ implementation to count of  ` `// palindromic Palindromic Substrings ` `// that can be formed from the array ` ` `  `#include  ` `using` `namespace` `std;  ` ` `  `// Function to to check if possible ` `// to make palindromic substring ` `bool` `isPossible(string A, string B)  ` `{  ` ` `  `        ``sort(B.begin(),B.end()); ` `        ``int` `c=0; ` `        ``for``(``int` `i = 0; i < (``int``)A.size(); i++) ` `            ``if``(binary_search(B.begin(),B.end(),A[i])) ` `                ``return` `true``; ` `    ``return` `false``; ` `}  ` ` `  `// Function to count of Palindromic Substrings ` `// that can be formed from the array. ` `int` `countPalindromicSubstrings(string s[], ``int` `n) ` `{ ` `    ``// variable to store count ` `    ``int` `count = 0; ` ` `  `    ``// Traverse through all the pairs ` `    ``// in the array ` `    ``for``(``int` `i = 0; i < n; i++){ ` `        ``for``(``int` `j = i + 1; j < n; j++) ` `            ``if``(isPossible(s[i], s[j])) ` `                ``count++; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver Code  ` `int` `main()  ` `{  ` `    ``string arr[] = { ``"gfg"``, ``"gfg"` `};  ` `    ``int` `n = 2; ` `    ``cout << countPalindromicSubstrings(arr, n); ` `    ``return` `0;  ` `}  `

## Java

 `// Java implementation to count of  ` `// palindromic Palindromic SubStrings ` `// that can be formed from the array ` `import` `java.util.*; ` ` `  `class` `GFG{  ` ` `  `// Function to to check if possible ` `// to make palindromic subString ` `static` `boolean` `isPossible(String A, String B)  ` `{  ` `    ``B = sortString(B); ` `     `  `    ``for``(``int` `i = ``0``; i < (``int``)A.length(); i++) ` `        ``if``(Arrays.binarySearch(B.toCharArray(),  ` `                               ``A.charAt(i)) > -``1``) ` `           ``return` `true``; ` `             `  `    ``return` `false``; ` `}  ` ` `  `// Function to count of Palindromic SubStrings ` `// that can be formed from the array. ` `static` `int` `countPalindromicSubStrings(String s[], ` `                                      ``int` `n) ` `{ ` `     `  `    ``// Variable to store count ` `    ``int` `count = ``0``; ` ` `  `    ``// Traverse through all the pairs ` `    ``// in the array ` `    ``for``(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``for``(``int` `j = i + ``1``; j < n; j++) ` `            ``if``(isPossible(s[i], s[j])) ` `                ``count++; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `static` `String sortString(String inputString)  ` `{  ` `     `  `    ``// Convert input string to char array  ` `    ``char` `tempArray[] = inputString.toCharArray();  ` `         `  `    ``// Sort tempArray  ` `    ``Arrays.sort(tempArray);  ` `         `  `    ``// Return new sorted string  ` `    ``return` `new` `String(tempArray);  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``String arr[] = { ``"gfg"``, ``"gfg"` `};  ` `    ``int` `n = ``2``; ` `     `  `    ``System.out.print(countPalindromicSubStrings(arr, n)); ` `}  ` `}  ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```1
```

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Improved By : Rajput-Ji