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Check if substrings from three given strings can be concatenated to form a palindrome
  • Last Updated : 02 Mar, 2021

Given three strings S1, S2, and S3 of lengths L, M, and N respectively, the task is to check if it is possible to choose some non-empty substrings from S1, S2, and S3 such that their concatenation is a palindrome. If found to be true, print “YES”. Otherwise, print “NO”.

Examples:

Input: S1 = “adcb”, S2 = “bcdb”, S3 = “dce”
Output: YES
Explanation: 
One of the possible ways is as follows:
Select substring “ad” from the string S1, “d” from the string S2, and “a” from the string S3. Therefore, the resultant string is S = “adda”, which is a palindrome. Therefore, the output should be “YES”.

Input: S1 = “a”, S2 = “bc”, S3 = “c”
Output: NO

Approach: The idea is to observe that it is always possible to find such a, b, and c such that a+b+c becomes palindrome if S1 and S3 have at least one common character. Follow the below steps to solve the problem:



  1. Initialize two variables, say maskA and maskC, for masking the characters in the strings S1 and S3 respectively.
  2. Iterate over the characters of the string S1 and set (i-‘a’)th bit in maskA, indicating that the respective character is present in string S1.
  3. Iterate over characters of the string S3 and set (i-‘a’)th bit in maskC, indicating that the respective character is present in string S3.
  4. If the Bitwise AND of maskA and maskC is greater than zero, then print “YES”. Otherwise, print “NO”.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if substrings from
// three given strings can be concatenated
// to form a palindrome
string make_palindrome(
    string S1, string S2, string S3)
{
    // Mask for S1 and S2
    int maskA = 0, maskC = 0;
 
    // Set (i-'a')th bit in maskA
    for (char i : S1)
        maskA |= (1 << (i - 'a'));
 
    // Set (i-'a')th bit in maskC
    for (char i : S3)
        maskC |= (1 << (i - 'a'));
 
    // If the bitwise AND is > 0
    if ((maskA & maskC) > 0)
        return "YES";
 
    return "NO";
}
 
// Driver Code
int main()
{
    string S1 = "adcb", S2 = "bcdb", S3 = "abe";
    cout << make_palindrome(S1, S2, S3);
}

Java




// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to check if subStrings from
// three given Strings can be concatenated
// to form a palindrome
static String make_palindrome(
    String S1, String S2, String S3)
{
   
    // Mask for S1 and S2
    int maskA = 0, maskC = 0;
 
    // Set (i-'a')th bit in maskA
    for (char i : S1.toCharArray())
        maskA |= (1 << (i - 'a'));
 
    // Set (i-'a')th bit in maskC
    for (char i : S3.toCharArray())
        maskC |= (1 << (i - 'a'));
 
    // If the bitwise AND is > 0
    if ((maskA & maskC) > 0)
        return "YES";
    return "NO";
}
 
// Driver Code
public static void main(String[] args)
{
    String S1 = "adcb", S2 = "bcdb", S3 = "abe";
    System.out.print(make_palindrome(S1, S2, S3));
}
}
 
// This code is contributed by shikhasingrajput

Python3




# Python3 program for the above approach
 
# Function to check if substrings from
# three given strings can be concatenated
# to form a palindrome
def make_palindrome(S1, S2, S3):
   
    # Mask for S1 and S2
    maskA, maskC = 0, 0
 
    # Set (i-'a')th bit in maskA
    for i in S1:
        maskA |= (1 << (ord(i) - ord('a')))
 
    # Set (i-'a')th bit in maskC
    for i in S3:
        maskC |= (1 << (ord(i) - ord('a')))
 
    # If the bitwise AND is > 0
    if ((maskA & maskC) > 0):
        return "YES"
 
    return "NO"
 
# Driver Code
if __name__ == '__main__':
    S1,S2,S3 = "adcb", "bcdb", "abe"
    print (make_palindrome(S1, S2, S3))
 
    # This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
public class GFG
{
 
// Function to check if subStrings from
// three given Strings can be concatenated
// to form a palindrome
static String make_palindrome(
    String S1, String S2, String S3)
{
   
    // Mask for S1 and S2
    int maskA = 0, maskC = 0;
 
    // Set (i-'a')th bit in maskA
    foreach (char i in S1.ToCharArray())
        maskA |= (1 << (i - 'a'));
 
    // Set (i-'a')th bit in maskC
    foreach (char i in S3.ToCharArray())
        maskC |= (1 << (i - 'a'));
 
    // If the bitwise AND is > 0
    if ((maskA & maskC) > 0)
        return "YES";
    return "NO";
}
 
// Driver Code
public static void Main(String[] args)
{
    String S1 = "adcb", S2 = "bcdb", S3 = "abe";
    Console.Write(make_palindrome(S1, S2, S3));
}
}
 
// This code is contributed by Rajput-Ji

 
 

Output: 
YES

 

Time Complexity: O(L + N)
Auxiliary Space: O(L + M + N) 

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