# Longest Palindrome in a String formed by concatenating its prefix and suffix

Last Updated : 26 Dec, 2022

Given a string str consisting of lowercase English letters, the task is to find the longest palindromic string T which satisfies the following condition:

• T = p + m + s where p and s are the prefix and the suffix of the given string str respectively and the string m is either the prefix or suffix of the string str after removing both p and s from it.
• The string formed by the concatenation of p and s is a palindrome itself.
• Either of the strings p and s can be an empty string.

Examples:

Input: str = “abcdfdcecba”
Output: abcdfdcba
Explanation:
Here, p = “abc”
s = “cba”
m = “dfd”
p + s = “abccba” which is a palindrome and m = “dfd” is the prefix after removing the prefix and suffix from the string str. Therefore, T = “abcdfdcba”.

Input: str = “geeksforgeeks”
Output:
Explanation:
Here, p = “”
s = “g”
m = “”
p + s = “” which is a palindrome and m = “g” is the prefix after removing the prefix and suffix from the string str. Therefore, T = “g”.

Approach: The idea for this problem is to divide the answer into three parts, such that there will be a part of suffix and prefix of the given string which forms palindrome together which will form the beginning and the ending of the answer string. Now, after removing these prefix and suffix from the given string, we can find the maximum length suffix or prefix string (which we may call midPalindrome) which is palindromic.
Therefore, the answer string will be given by:

`answer = prefix + midPalindrome + suffix`

The following steps can be followed to compute the answer to the problem:

• Find the length up to which the suffix and prefix of str form a palindrome together.
• Remove the suffix and prefix substrings that already form a palindrome from str and store them in separate strings.
• Check all prefix and suffix substrings in the remaining string str and find the longest of such strings.
• Finally, combine all three parts of the answer and return it.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the longest ` `// palindrome in a string formed by ` `// concatenating its prefix and suffix`   `#include ` `using` `namespace` `std;`   `// Function to check whether` `// the string is a palindrome` `bool` `isPalindrome(string r)` `{` `    ``string p = r;`   `    ``// Reverse the string to` `    ``// compare with the` `    ``// original string` `    ``reverse(p.begin(), p.end());`   `    ``// Check if both are same` `    ``return` `(r == p);` `}`   `// Function to find the longest` `// palindrome in a string formed by` `// concatenating its prefix and suffix` `string PrefixSuffixPalindrome(string str)` `{` `    ``// Length of the string` `    ``int` `n = str.size(), len = 0;`   `    ``// Finding the length upto which` `    ``// the suffix and prefix forms a` `    ``// palindrome together` `    ``for` `(``int` `i = 0; i < n / 2; i++) {` `        ``if` `(str[i] != str[n - i - 1]) {` `            ``len = i;` `            ``break``;` `        ``}` `    ``}`   `    ``// Check whether the string` `    ``// has prefix and suffix substrings` `    ``// which are palindromes.` `    ``string prefix = ``""``, suffix = ``""``;` `    ``string midPal = ``""``;`   `    ``// Removing the suffix and prefix` `    ``// substrings which already forms` `    ``// a palindrome and storing them` `    ``// in separate strings` `    ``prefix = str.substr(0, len);` `    ``suffix = str.substr(n - len);` `    ``str = str.substr(len, n - 2 * len);`   `    ``// Check all prefix substrings` `    ``// in the remaining string str` `    ``for` `(``int` `i = 1; i <= str.size(); i++) {` `        ``string y = str.substr(0, i);`   `        ``// Check if the prefix substring` `        ``// is a palindrome` `        ``if` `(isPalindrome(y)) {`   `            ``// If the prefix substring` `            ``// is a palindrome then check` `            ``// if it is of maximum length` `            ``// so far` `            ``if` `(midPal.size() < y.size()) {` `                ``midPal = y;` `            ``}` `        ``}` `    ``}`   `    ``// Check all the suffix substrings` `    ``// in the remaining string str` `    ``for` `(``int` `i = 1; i <= str.size(); i++) {` `        ``string y = str.substr(str.size() - i);`   `        ``// Check if the suffix substring` `        ``// is a palindrome` `        ``if` `(isPalindrome(y)) {`   `            ``// If the suffix substring` `            ``// is a palindrome then check` `            ``// if it is of maximum length` `            ``// so far` `            ``if` `(midPal.size() < y.size()) {` `                ``midPal = y;` `            ``}` `        ``}` `    ``}`   `    ``// Combining all the three parts` `    ``// of the answer` `    ``string answer = prefix + midPal + suffix;`   `    ``return` `answer;` `}`   `// Driver code` `int` `main()` `{` `    ``string str = ``"abcdfdcecba"``;`   `    ``cout << PrefixSuffixPalindrome(str) << ``"\n"``;`   `    ``return` `0;` `}`

## Java

 `// Java program to find the longest ` `// palindrome in a String formed by ` `// concatenating its prefix and suffix` `import` `java.util.*;`   `class` `GFG{` ` `  `// Function to check whether` `// the String is a palindrome` `static` `boolean` `isPalindrome(String r)` `{` `    ``String p = r;` ` `  `    ``// Reverse the String to` `    ``// compare with the` `    ``// original String` `    ``p = reverse(p);` ` `  `    ``// Check if both are same` `    ``return` `(r.equals(p));` `}` ` `  `// Function to find the longest` `// palindrome in a String formed by` `// concatenating its prefix and suffix` `static` `String PrefixSuffixPalindrome(String str)` `{` `    ``// Length of the String` `    ``int` `n = str.length(), len = ``0``;` ` `  `    ``// Finding the length upto which` `    ``// the suffix and prefix forms a` `    ``// palindrome together` `    ``for` `(``int` `i = ``0``; i < n / ``2``; i++) {` `        ``if` `(str.charAt(i) != str.charAt(n - i - ``1``)) {` `            ``len = i;` `            ``break``;` `        ``}` `    ``}` ` `  `    ``// Check whether the String` `    ``// has prefix and suffix subStrings` `    ``// which are palindromes.` `    ``String prefix = ``""``, suffix = ``""``;` `    ``String midPal = ``""``;` ` `  `    ``// Removing the suffix and prefix` `    ``// subStrings which already forms` `    ``// a palindrome and storing them` `    ``// in separate Strings` `    ``prefix = str.substring(``0``, len);` `    ``suffix = str.substring(n - len);` `    ``str = str.substring(len, (n - ``2` `* len) + len);` ` `  `    ``// Check all prefix subStrings` `    ``// in the remaining String str` `    ``for` `(``int` `i = ``1``; i <= str.length(); i++) {` `        ``String y = str.substring(``0``, i);` ` `  `        ``// Check if the prefix subString` `        ``// is a palindrome` `        ``if` `(isPalindrome(y)) {` ` `  `            ``// If the prefix subString` `            ``// is a palindrome then check` `            ``// if it is of maximum length` `            ``// so far` `            ``if` `(midPal.length() < y.length()) {` `                ``midPal = y;` `            ``}` `        ``}` `    ``}` ` `  `    ``// Check all the suffix subStrings` `    ``// in the remaining String str` `    ``for` `(``int` `i = ``1``; i <= str.length(); i++) {` `        ``String y = str.substring(str.length() - i);` ` `  `        ``// Check if the suffix subString` `        ``// is a palindrome` `        ``if` `(isPalindrome(y)) {` ` `  `            ``// If the suffix subString` `            ``// is a palindrome then check` `            ``// if it is of maximum length` `            ``// so far` `            ``if` `(midPal.length() < y.length()) {` `                ``midPal = y;` `            ``}` `        ``}` `    ``}` ` `  `    ``// Combining all the parts` `    ``// of the answer` `    ``String answer = prefix + midPal + suffix;` ` `  `    ``return` `answer;` `}` `static` `String reverse(String input) {` `    ``char``[] a = input.toCharArray();` `    ``int` `l, r = a.length - ``1``;` `    ``for` `(l = ``0``; l < r; l++, r--) {` `        ``char` `temp = a[l];` `        ``a[l] = a[r];` `        ``a[r] = temp;` `    ``}` `    ``return` `String.valueOf(a);` `}` ` `  `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``String str = ``"abcdfdcecba"``;` ` `  `    ``System.out.print(PrefixSuffixPalindrome(str));` `}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to find the longest ` `# palindrome in a string formed by ` `# concatenating its prefix and suffix` ` `  `# Function to check whether` `# the string is a palindrome` `def` `isPalindrome(r):`   `    ``p ``=` `r` ` `  `    ``# Reverse the string to` `    ``# compare with the` `    ``# original string` `    ``p ``=` `"".join(``reversed``(p))` ` `  `    ``# Check if both are same` `    ``return` `(r ``=``=` `p)` ` `  `# Function to find the longest` `# palindrome in a string formed by` `# concatenating its prefix and suffix` `def` `PrefixSuffixPalindrome(st):`   `    ``# Length of the string` `    ``n ``=` `len``(st)` `    ``length ``=` `0` ` `  `    ``# Finding the length upto which` `    ``# the suffix and prefix forms a` `    ``# palindrome together` `    ``for` `i ``in` `range``( n ``/``/` `2``):` `        ``if` `(st[i] !``=` `st[n ``-` `i ``-` `1``]):` `            ``length ``=` `i` `            ``break` ` `  `    ``# Check whether the string` `    ``# has prefix and suffix substrings` `    ``# which are palindromes.` `    ``prefix ``=` `""` `    ``suffix ``=` `""` `    ``midPal ``=` `""` ` `  `    ``# Removing the suffix and prefix` `    ``# substrings which already forms` `    ``# a palindrome and storing them` `    ``# in separate strings` `    ``prefix ``=` `st[:length]` `    ``suffix ``=` `st[n ``-` `length:]` `    ``st ``=` `st[length: n ``-` `2` `*` `length``+``length]`   `    ``# Check all prefix substrings` `    ``# in the remaining string str` `    ``for` `i ``in` `range``(``1``,``len``(st)``+``1``):` `        ``y ``=` `st[``0``: i]` ` `  `        ``# Check if the prefix substring` `        ``# is a palindrome` `        ``if` `(isPalindrome(y)):` ` `  `            ``# If the prefix substring` `            ``# is a palindrome then check` `            ``# if it is of maximum length` `            ``# so far` `            ``if` `(``len``(midPal) < ``len``(y)):` `                ``midPal ``=` `y` ` `  `    ``# Check all the suffix substrings` `    ``# in the remaining string str` `    ``for` `i ``in` `range``(``1``,``len``(st)``+``1``):` `        ``y ``=` `st[``len``(st)``-``i]` ` `  `        ``# Check if the suffix substring` `        ``# is a palindrome` `        ``if` `(isPalindrome(y)):` ` `  `            ``# If the suffix substring` `            ``# is a palindrome then check` `            ``# if it is of maximum length` `            ``# so far` `            ``if` `(``len``(midPal) < ``len``(y)):` `                ``midPal ``=` `y` ` `  `    ``# Combining all the parts` `    ``# of the answer` `    ``answer ``=` `prefix ``+` `midPal ``+` `suffix` ` `  `    ``return` `answer` ` `  `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` `    `  `    ``st ``=` `"abcdfdcecba"``;` ` `  `    ``print``(PrefixSuffixPalindrome(st))` ` `  `# This code is contributed by chitranayal` `   `

## C#

 `// C# program to find the longest ` `// palindrome in a String formed by ` `// concatenating its prefix and suffix` `using` `System;`   `class` `GFG{` `  `  `// Function to check whether` `// the String is a palindrome` `static` `bool` `isPalindrome(String r)` `{` `    ``String p = r;` `  `  `    ``// Reverse the String to` `    ``// compare with the` `    ``// original String` `    ``p = reverse(p);` `  `  `    ``// Check if both are same` `    ``return` `(r.Equals(p));` `}` `  `  `// Function to find the longest` `// palindrome in a String formed by` `// concatenating its prefix and suffix` `static` `String PrefixSuffixPalindrome(String str)` `{` `    ``// Length of the String` `    ``int` `n = str.Length, len = 0;` `  `  `    ``// Finding the length upto which` `    ``// the suffix and prefix forms a` `    ``// palindrome together` `    ``for` `(``int` `i = 0; i < n / 2; i++) {` `        ``if` `(str[i] != str[n - i - 1]) {` `            ``len = i;` `            ``break``;` `        ``}` `    ``}` `  `  `    ``// Check whether the String` `    ``// has prefix and suffix subStrings` `    ``// which are palindromes.` `    ``String prefix = ``""``, suffix = ``""``;` `    ``String midPal = ``""``;` `  `  `    ``// Removing the suffix and prefix` `    ``// subStrings which already forms` `    ``// a palindrome and storing them` `    ``// in separate Strings` `    ``prefix = str.Substring(0, len);` `    ``suffix = str.Substring(n - len);` `    ``str = str.Substring(len, (n - 2 * len) + len);` `  `  `    ``// Check all prefix subStrings` `    ``// in the remaining String str` `    ``for` `(``int` `i = 1; i <= str.Length; i++) {` `        ``String y = str.Substring(0, i);` `  `  `        ``// Check if the prefix subString` `        ``// is a palindrome` `        ``if` `(isPalindrome(y)) {` `  `  `            ``// If the prefix subString` `            ``// is a palindrome then check` `            ``// if it is of maximum length` `            ``// so far` `            ``if` `(midPal.Length < y.Length) {` `                ``midPal = y;` `            ``}` `        ``}` `    ``}` `  `  `    ``// Check all the suffix subStrings` `    ``// in the remaining String str` `    ``for` `(``int` `i = 1; i <= str.Length; i++) {` `        ``String y = str.Substring(str.Length - i);` `  `  `        ``// Check if the suffix subString` `        ``// is a palindrome` `        ``if` `(isPalindrome(y)) {` `  `  `            ``// If the suffix subString` `            ``// is a palindrome then check` `            ``// if it is of maximum length` `            ``// so far` `            ``if` `(midPal.Length < y.Length) {` `                ``midPal = y;` `            ``}` `        ``}` `    ``}` `  `  `    ``// Combining all the parts` `    ``// of the answer` `    ``String answer = prefix + midPal + suffix;` `  `  `    ``return` `answer;` `}` `static` `String reverse(String input) {` `    ``char``[] a = input.ToCharArray();` `    ``int` `l, r = a.Length - 1;` `    ``for` `(l = 0; l < r; l++, r--) {` `        ``char` `temp = a[l];` `        ``a[l] = a[r];` `        ``a[r] = temp;` `    ``}` `    ``return` `String.Join(``""``,a);` `}` `  `  `// Driver code` `public` `static` `void` `Main(String[] args)` `{` `    ``String str = ``"abcdfdcecba"``;` `  `  `    ``Console.Write(PrefixSuffixPalindrome(str));` `}` `}` ` `  `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`abcdfdcba`

Time Complexity: O(n2)
Auxiliary Space: O(n), where n is the length of the given string.

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