# Check if a Palindromic String can be formed by concatenating Substrings of two given Strings

Given two strings str1 and str2, the task is to check if it is possible to form a Palindromic String by concatenation of two substrings of str1 and str2.

Examples:

Input: str1 = “abcd”, str2 = “acba”
Output: Yes
Explanation:
There are five possible cases where concatenation of two substrings from str1 and str2 gives palindromic string:
“ab” + “a” = “aba”
“ab” + “ba” = “abba”
“bc” + “cb” = “bccb”
“bc” + “b” = “bcb”
“cd” + “c” = “cdc”

Input: str1 = “pqrs”, str2 = “abcd”
Output: No
Explanation:
There is no possible concatenation of sub-strings from given strings which gives palindromic string.

Naive Approach:
The simplest approach to solve the problem is to generate every possible substring of str1 and str2 and combine them to generate all possible concatenations. For each concatenation, check if it is palindromic or not. If found to be true, print “Yes”. Otherwise, print “No”.

Time Complexity: O(N2* M2 * (N+M)), where N and M are the lengths of str1 and str2 respectively.
Auxiliary Space: O(1)

Efficient Approach:
To optimize the above approach, the following observation needs to be made:

If the given strings possess at least one common character, then they will always form a palindromic string on concatenation of the common character from both the strings.
Illustration:
str1 = “abc”, str2 = “fad”
Since ‘a’ is common in both strings, a palindromic string “aa” can be obtained.

Follow the steps below to solve the problem:

• Initialize a boolean array to mark the presence of each alphabet in the two strings.
• Traverse str1 and mark the index (str1[i] – ‘a’) as true.
• Now, traverse str2 and check if any index (str2[i] – ‘a’) is already marked as true, print “Yes”.
• After complete traversal of str2, if no common character is found, print “No”.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement  ` `// the above approach  ` `#include   ` `using` `namespace` `std;  ` ` `  `// Function to check if a palindromic  ` `// string can be formed from the  ` `// substring of given strings  ` `bool` `check(string str1, string str2)  ` `{  ` `    ``// Boolean array to mark  ` `    ``// presence of characters  ` `    ``vector<``bool``> mark(26, ``false``);  ` ` `  `    ``int` `n = str1.size(),  ` `        ``m = str2.size();  ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) {  ` ` `  `        ``mark[str1[i] - ``'a'``] = ``true``;  ` `    ``}  ` ` `  `    ``// Check if any of the character  ` `    ``// of str2 is already marked  ` `    ``for` `(``int` `i = 0; i < m; i++) {  ` ` `  `        ``// If a common character  ` `        ``// is found  ` `        ``if` `(mark[str2[i] - ``'a'``])  ` `            ``return` `true``;  ` `    ``}  ` ` `  `    ``// If no common character  ` `    ``// is found  ` `    ``return` `false``;  ` `}  ` ` `  `// Driver Code  ` `int` `main()  ` `{  ` ` `  `    ``string str1 = ``"abca"``,  ` `        ``str2 = ``"efad"``;  ` ` `  `    ``if` `(check(str1, str2))  ` `        ``cout << ``"Yes"``;  ` `    ``else` `        ``cout << ``"No"``;  ` ` `  `    ``return` `0;  ` `}  `

## Java

 `// Java program to implement  ` `// the above approach  ` `import` `java.util.Arrays;  ` ` `  `class` `GFG{  ` `     `  `// Function to check if a palindromic  ` `// string can be formed from the  ` `// substring of given strings  ` `public` `static` `boolean` `check(String str1,  ` `                            ``String str2)  ` `{  ` `     `  `    ``// Boolean array to mark  ` `    ``// presence of characters  ` `    ``boolean``[] mark = ``new` `boolean``[``26``];  ` `    ``Arrays.fill(mark, ``false``);  ` `     `  `    ``int` `n = str1.length(),  ` `        ``m = str2.length();  ` ` `  `    ``for``(``int` `i = ``0``; i < n; i++)  ` `    ``{  ` `        ``mark[str1.charAt(i) - ``'a'``] = ``true``;  ` `    ``}  ` ` `  `    ``// Check if any of the character  ` `    ``// of str2 is already marked  ` `    ``for``(``int` `i = ``0``; i < m; i++)  ` `    ``{  ` ` `  `        ``// If a common character  ` `        ``// is found  ` `        ``if` `(mark[str2.charAt(i) - ``'a'``])  ` `            ``return` `true``;  ` `    ``}  ` ` `  `    ``// If no common character  ` `    ``// is found  ` `    ``return` `false``;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``String str1 = ``"abca"``,  ` `    ``str2 = ``"efad"``;  ` ` `  `    ``if` `(check(str1, str2))  ` `        ``System.out.println(``"Yes"``);  ` `    ``else` `        ``System.out.println(``"No"``);  ` `}  ` `}  ` ` `  `// This code is contributed by divyeshrabadiya07  `

## Python3

 `# Python3 program to implement  ` `# the above approach  ` `     `  `# Function to check if a palindromic  ` `# string can be formed from the  ` `# substring of given strings  ` `def` `check(str1, str2): ` `     `  `    ``# Boolean array to mark  ` `    ``# presence of characters  ` `    ``mark ``=` `[``False` `for` `i ``in` `range``(``26``)] ` `     `  `    ``n ``=` `len``(str1) ` `    ``m ``=` `len``(str2) ` `     `  `    ``for` `i ``in` `range``(n): ` `        ``mark[``ord``(str1[i]) ``-` `ord``(``'a'``)] ``=` `True` `     `  `    ``# Check if any of the character  ` `    ``# of str2 is already marked  ` `    ``for` `i ``in` `range``(m): ` `         `  `        ``# If a common character  ` `        ``# is found  ` `        ``if` `(mark[``ord``(str2[i]) ``-` `ord``(``'a'``)]): ` `            ``return` `True``;  ` ` `  `    ``# If no common character  ` `    ``# is found  ` `    ``return` `False` `     `  `# Driver code ` `if` `__name__``=``=``"__main__"``: ` `     `  `    ``str1 ``=` `"abca"` `    ``str2 ``=` `"efad"` ` `  `    ``if` `(check(str1, str2)): ` `        ``print``(``"Yes"``); ` `    ``else``: ` `        ``print``(``"No"``); ` ` `  `# This code is contributed by rutvik_56 `

## C#

 `// C# program to implement  ` `// the above approach  ` `using` `System;  ` ` `  `class` `GFG{  ` `     `  `// Function to check if a palindromic  ` `// string can be formed from the  ` `// substring of given strings  ` `public` `static` `bool` `check(String str1,  ` `                        ``String str2)  ` `{  ` `     `  `    ``// Boolean array to mark  ` `    ``// presence of characters  ` `    ``bool``[] mark = ``new` `bool``[26];  ` `     `  `    ``int` `n = str1.Length,  ` `        ``m = str2.Length;  ` ` `  `    ``for``(``int` `i = 0; i < n; i++)  ` `    ``{  ` `        ``mark[str1[i] - ``'a'``] = ``true``;  ` `    ``}  ` ` `  `    ``// Check if any of the character  ` `    ``// of str2 is already marked  ` `    ``for``(``int` `i = 0; i < m; i++)  ` `    ``{  ` ` `  `        ``// If a common character  ` `        ``// is found  ` `        ``if` `(mark[str2[i] - ``'a'``])  ` `            ``return` `true``;  ` `    ``}  ` ` `  `    ``// If no common character  ` `    ``// is found  ` `    ``return` `false``;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main(String[] args)  ` `{  ` `    ``String str1 = ``"abca"``,  ` `    ``str2 = ``"efad"``;  ` ` `  `    ``if` `(check(str1, str2))  ` `        ``Console.WriteLine(``"Yes"``);  ` `    ``else` `        ``Console.WriteLine(``"No"``);  ` `}  ` `}  ` ` `  `// This code is contributed by amal kumar choubey  `

Output:

```Yes
```

Time Complexity: O(max(N, M)) where N and M are the lengths of str1 and str2 respectively.
Auxiliary Space: O(1)

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