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Count pair of strings whose concatenation of substrings form a palindrome

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  • Difficulty Level : Basic
  • Last Updated : 09 Jun, 2022

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<bits/stdc++.h>
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

Python3




# Python3 implementation to count of
# palindromic Palindromic Substrings
# that can be formed from the array
 
# Function to to check if possible
# to make palindromic substring
def isPossible(A, B):
   
    B = sorted(B)
    c = 0
     
    for i in range(len(A)):
        if A[i] in B:
            return True
    return False
 
# Function to count of Palindromic
# Substrings that can be formed
# from the array.
def countPalindromicSubstrings(s, n):
 
    # Variable to store count
    count = 0
 
    # Traverse through all
    # Substrings in the array
    for i in range(n):
        for j in range(i + 1, n):
            if(isPossible(s[i], s[j])):
                count += 1
    return count
 
# Driver Code
arr = ["gfg", "gfg"]
n = 2
print(countPalindromicSubstrings(arr, n))
 
# This code is contributed by avanitrachhadiya2155

C#




// C# implementation to count of
// palindromic Palindromic SubStrings
// that can be formed from the array
using System;
class GFG{
 
// Function to to check if possible
// to make palindromic subString
static bool isPossible(String A, String B)
{
    B = sortString(B);
     
    for(int i = 0; i < (int)A.Length; i++)
        if(Array.BinarySearch(B.ToCharArray(),
                               A[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
    Array.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;
     
    Console.Write(countPalindromicSubStrings(arr, n));
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
// Javascript implementation
 
// Function to to check if possible
// to make palindromic substring
function isPossible(A, B)
{
    B = B.split('').sort().join('');
    var c=0;
    for(var i = 0; i < A.length; i++)
        if(B.indexOf(A[i]) != -1)
            return true;
    return false;
}
   
// Function to count of Palindromic Substrings
// that can be formed from the array.
function countPalindromicSubstrings(s, n)
{
    // variable to store count
    var count = 0;
   
    // Traverse through all the pairs
    // in the array
    for(var i = 0; i < n; i++){
        for(var j = i + 1; j < n; j++)
            if(isPossible(s[i], s[j]))
                count++;
    }
    return count;
}
 
// Driver Code
var arr = ["gfg", "gfg"]
var n = 2
document.write(countPalindromicSubstrings(arr, n))
   
// This code is contributed by shubhamsingh10
</script>

Output: 

1

 

Time complexity: O(n2*mlogm) where m is length of string

Auxiliary Space: O(1)


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