# Longest Palindromic Substring | Set 1

• Difficulty Level : Hard
• Last Updated : 16 Jun, 2022

Given a string, find the longest substring which is palindrome.

For example,

```Input: Given string :"forgeeksskeegfor",
Output: "geeksskeeg"

Input: Given string :"Geeks",
Output: "ee"```

Method 1: Brute Force.
Approach: The simple approach is to check each substring whether the substring is a palindrome or not. To do this first, run three nested loops, the outer two loops pick all substrings one by one by fixing the corner characters, the inner loop checks whether the picked substring is palindrome or not.

## C++

 `// A C++ solution for longest palindrome` `#include ` `using` `namespace` `std;`   `// Function to print a substring str[low..high]` `void` `printSubStr(string str, ``int` `low, ``int` `high)` `{` `    ``for` `(``int` `i = low; i <= high; ++i)` `        ``cout << str[i];` `}`   `// This function prints the` `// longest palindrome substring` `// It also returns the length` `// of the longest palindrome` `int` `longestPalSubstr(string str)` `{` `    ``// get length of input string` `    ``int` `n = str.size();`   `    ``// All substrings of length 1` `    ``// are palindromes` `    ``int` `maxLength = 1, start = 0;`   `    ``// Nested loop to mark start and end index` `    ``for` `(``int` `i = 0; i < str.length(); i++) {` `        ``for` `(``int` `j = i; j < str.length(); j++) {` `            ``int` `flag = 1;`   `            ``// Check palindrome` `            ``for` `(``int` `k = 0; k < (j - i + 1) / 2; k++)` `                ``if` `(str[i + k] != str[j - k])` `                    ``flag = 0;`   `            ``// Palindrome` `            ``if` `(flag && (j - i + 1) > maxLength) {` `                ``start = i;` `                ``maxLength = j - i + 1;` `            ``}` `        ``}` `    ``}`   `    ``cout << ``"Longest palindrome substring is: "``;` `    ``printSubStr(str, start, start + maxLength - 1);`   `    ``// return length of LPS` `    ``return` `maxLength;` `}`   `// Driver Code` `int` `main()` `{` `    ``string str = ``"forgeeksskeegfor"``;` `    ``cout << ``"\nLength is: "` `         ``<< longestPalSubstr(str);` `    ``return` `0;` `}`

## Java

 `// A Java solution for longest palindrome` `import` `java.util.*;`   `class` `GFG{`   `// Function to print a subString str[low..high]` `static` `void` `printSubStr(String str, ``int` `low, ``int` `high)` `{` `    ``for` `(``int` `i = low; i <= high; ++i)` `        ``System.out.print(str.charAt(i));` `}`   `// This function prints the` `// longest palindrome subString` `// It also returns the length` `// of the longest palindrome` `static` `int` `longestPalSubstr(String str)` `{` `    ``// get length of input String` `    ``int` `n = str.length();`   `    ``// All subStrings of length 1` `    ``// are palindromes` `    ``int` `maxLength = ``1``, start = ``0``;`   `    ``// Nested loop to mark start and end index` `    ``for` `(``int` `i = ``0``; i < str.length(); i++) {` `        ``for` `(``int` `j = i; j < str.length(); j++) {` `            ``int` `flag = ``1``;`   `            ``// Check palindrome` `            ``for` `(``int` `k = ``0``; k < (j - i + ``1``) / ``2``; k++)` `                ``if` `(str.charAt(i + k) != str.charAt(j - k))` `                    ``flag = ``0``;`   `            ``// Palindrome` `            ``if` `(flag!=``0` `&& (j - i + ``1``) > maxLength) {` `                ``start = i;` `                ``maxLength = j - i + ``1``;` `            ``}` `        ``}` `    ``}`   `    ``System.out.print(``"Longest palindrome subString is: "``);` `    ``printSubStr(str, start, start + maxLength - ``1``);`   `    ``// return length of LPS` `    ``return` `maxLength;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``String str = ``"forgeeksskeegfor"``;` `    ``System.out.print(``"\nLength is: "` `         ``+ longestPalSubstr(str));` `}` `}`   `// This code is contributed by shikhasingrajput`

## Python3

 `# A Python3 solution for longest palindrome`   `# Function to pra subString str[low..high]` `def` `printSubStr(``str``, low, high):` `    `  `    ``for` `i ``in` `range``(low, high ``+` `1``):` `        ``print``(``str``[i], end ``=` `"")`   `# This function prints the` `# longest palindrome subString` `# It also returns the length` `# of the longest palindrome` `def` `longestPalSubstr(``str``):` `    `  `    ``# Get length of input String` `    ``n ``=` `len``(``str``)` `    `  `    ``# All subStrings of length 1` `    ``# are palindromes` `    ``maxLength ``=` `1` `    ``start ``=` `0` `    `  `    ``# Nested loop to mark start` `    ``# and end index` `    ``for` `i ``in` `range``(n):` `        ``for` `j ``in` `range``(i, n):` `            ``flag ``=` `1` `            `  `            ``# Check palindrome` `            ``for` `k ``in` `range``(``0``, ((j ``-` `i) ``/``/` `2``) ``+` `1``):` `                ``if` `(``str``[i ``+` `k] !``=` `str``[j ``-` `k]):` `                    ``flag ``=` `0`   `            ``# Palindrome` `            ``if` `(flag !``=` `0` `and` `(j ``-` `i ``+` `1``) > maxLength):` `                ``start ``=` `i` `                ``maxLength ``=` `j ``-` `i ``+` `1` `                `  `    ``print``(``"Longest palindrome subString is: "``, end ``=` `"")` `    ``printSubStr(``str``, start, start ``+` `maxLength ``-` `1``)`   `    ``# Return length of LPS` `    ``return` `maxLength`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``str` `=` `"forgeeksskeegfor"` `    `  `    ``print``(``"\nLength is: "``, longestPalSubstr(``str``))`   `# This code is contributed by 29AjayKumar`

## C#

 `// A C# solution for longest palindrome` `using` `System;`   `class` `GFG{`   `// Function to print a subString str[low..high]` `static` `void` `printSubStr(String str, ``int` `low, ``int` `high)` `{` `    ``for` `(``int` `i = low; i <= high; ++i)` `        ``Console.Write(str[i]);` `}`   `// This function prints the` `// longest palindrome subString` `// It also returns the length` `// of the longest palindrome` `static` `int` `longestPalSubstr(String str)` `{` `    ``// get length of input String` `    ``int` `n = str.Length;`   `    ``// All subStrings of length 1` `    ``// are palindromes` `    ``int` `maxLength = 1, start = 0;`   `    ``// Nested loop to mark start and end index` `    ``for` `(``int` `i = 0; i < str.Length; i++) {` `        ``for` `(``int` `j = i; j < str.Length; j++) {` `            ``int` `flag = 1;`   `            ``// Check palindrome` `            ``for` `(``int` `k = 0; k < (j - i + 1) / 2; k++)` `                ``if` `(str[i + k] != str[j - k])` `                    ``flag = 0;`   `            ``// Palindrome` `            ``if` `(flag!=0 && (j - i + 1) > maxLength) {` `                ``start = i;` `                ``maxLength = j - i + 1;` `            ``}` `        ``}` `    ``}`   `    ``Console.Write(``"longest palindrome subString is: "``);` `    ``printSubStr(str, start, start + maxLength - 1);`   `    ``// return length of LPS` `    ``return` `maxLength;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``String str = ``"forgeeksskeegfor"``;` `    ``Console.Write(``"\nLength is: "` `         ``+ longestPalSubstr(str));` `}` `}`   `// This code is contributed by shikhasingrajput`

## Javascript

 ``

Output

```Longest palindrome substring is: geeksskeeg
Length is: 10```

Complexity Analysis:

• Time complexity: O(n^3).
Three nested loops are needed to find the longest palindromic substring in this approach, so the time complexity is O(n^3).
• Auxiliary complexity: O(1).
As no extra space is needed.

Method 2: Dynamic Programming.
Approach: The time complexity can be reduced by storing results of sub-problems. The idea is similar to this post.

1. Maintain a boolean table[n][n] that is filled in bottom up manner.
2. The value of table[i][j] is true, if the substring is palindrome, otherwise false.
3. To calculate table[i][j], check the value of table[i+1][j-1], if the value is true and str[i] is same as str[j], then we make table[i][j] true.
4. Otherwise, the value of table[i][j] is made false.
5. We have to fill table previously for substring of length = 1 and length =2 because
as we are finding , if table[i+1][j-1] is true or false , so in case of
(i) length == 1 , lets say i=2 , j=2 and i+1,j-1 doesn’t lies between [i , j]
(ii) length == 2 ,lets say i=2 , j=3 and i+1,j-1 again doesn’t lies between [i , j].

Below is the implementation of the above approach:

## C++

 `// A C++ dynamic programming` `// solution for longest palindrome`   `#include ` `using` `namespace` `std;`   `// Function to print a substring` `// str[low..high]` `void` `printSubStr(` `    ``string str, ``int` `low, ``int` `high)` `{` `    ``for` `(``int` `i = low; i <= high; ++i)` `        ``cout << str[i];` `}`   `// This function prints the` `// longest palindrome substring` `// It also returns the length of` `// the longest palindrome` `int` `longestPalSubstr(string str)` `{` `    ``// get length of input string` `    ``int` `n = str.size();`   `    ``// table[i][j] will be false if substring` `    ``// str[i..j] is not palindrome.` `    ``// Else table[i][j] will be true` `    ``bool` `table[n][n];`   `    ``memset``(table, 0, ``sizeof``(table));`   `    ``// All substrings of length 1` `    ``// are palindromes` `    ``int` `maxLength = 1;`   `    ``for` `(``int` `i = 0; i < n; ++i)` `        ``table[i][i] = ``true``;`   `    ``// check for sub-string of length 2.` `    ``int` `start = 0;` `    ``for` `(``int` `i = 0; i < n - 1; ++i) {` `        ``if` `(str[i] == str[i + 1]) {` `            ``table[i][i + 1] = ``true``;` `            ``start = i;` `            ``maxLength = 2;` `        ``}` `    ``}`   `    ``// Check for lengths greater than 2.` `    ``// k is length of substring` `    ``for` `(``int` `k = 3; k <= n; ++k) {` `        ``// Fix the starting index` `        ``for` `(``int` `i = 0; i < n - k + 1; ++i) {` `            ``// Get the ending index of substring from` `            ``// starting index i and length k` `            ``int` `j = i + k - 1;`   `            ``// checking for sub-string from ith index to` `            ``// jth index iff str[i+1] to str[j-1] is a` `            ``// palindrome` `            ``if` `(table[i + 1][j - 1] && str[i] == str[j]) {` `                ``table[i][j] = ``true``;`   `                ``if` `(k > maxLength) {` `                    ``start = i;` `                    ``maxLength = k;` `                ``}` `            ``}` `        ``}` `    ``}`   `    ``cout << ``"Longest palindrome substring is: "``;` `    ``printSubStr(str, start, start + maxLength - 1);`   `    ``// return length of LPS` `    ``return` `maxLength;` `}`   `// Driver Code` `int` `main()` `{` `    ``string str = ``"forgeeksskeegfor"``;` `    ``cout << ``"\nLength is: "` `         ``<< longestPalSubstr(str);` `    ``return` `0;` `}`

## Java

 `// Java Solution` `public` `class` `LongestPalinSubstring {` `    ``// A utility function to print` `    ``// a substring str[low..high]` `    ``static` `void` `printSubStr(` `        ``String str, ``int` `low, ``int` `high)` `    ``{` `        ``System.out.println(` `            ``str.substring(` `                ``low, high + ``1``));` `    ``}`   `    ``// This function prints the longest` `    ``// palindrome substring of str[].` `    ``// It also returns the length of the` `    ``// longest palindrome` `    ``static` `int` `longestPalSubstr(String str)` `    ``{` `        ``// get length of input string` `        ``int` `n = str.length();`   `        ``// table[i][j] will be false if` `        ``// substring str[i..j] is not palindrome.` `        ``// Else table[i][j] will be true` `        ``boolean` `table[][] = ``new` `boolean``[n][n];`   `        ``// All substrings of length 1 are palindromes` `        ``int` `maxLength = ``1``;` `        ``for` `(``int` `i = ``0``; i < n; ++i)` `            ``table[i][i] = ``true``;`   `        ``// check for sub-string of length 2.` `        ``int` `start = ``0``;` `        ``for` `(``int` `i = ``0``; i < n - ``1``; ++i) {` `            ``if` `(str.charAt(i) == str.charAt(i + ``1``)) {` `                ``table[i][i + ``1``] = ``true``;` `                ``start = i;` `                ``maxLength = ``2``;` `            ``}` `        ``}`   `        ``// Check for lengths greater than 2.` `        ``// k is length of substring` `        ``for` `(``int` `k = ``3``; k <= n; ++k) {`   `            ``// Fix the starting index` `            ``for` `(``int` `i = ``0``; i < n - k + ``1``; ++i) {` `                ``// Get the ending index of substring from` `                ``// starting index i and length k` `                ``int` `j = i + k - ``1``;`   `                ``// checking for sub-string from ith index to` `                ``// jth index iff str.charAt(i+1) to` `                ``// str.charAt(j-1) is a palindrome` `                ``if` `(table[i + ``1``][j - ``1``]` `                    ``&& str.charAt(i) == str.charAt(j)) {` `                    ``table[i][j] = ``true``;`   `                    ``if` `(k > maxLength) {` `                        ``start = i;` `                        ``maxLength = k;` `                    ``}` `                ``}` `            ``}` `        ``}` `        ``System.out.print(``"Longest palindrome substring is; "``);` `        ``printSubStr(str, start,` `                    ``start + maxLength - ``1``);`   `        ``// return length of LPS` `        ``return` `maxLength;` `    ``}`   `    ``// Driver program to test above functions` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``String str = ``"forgeeksskeegfor"``;` `        ``System.out.println(``"Length is: "` `+ longestPalSubstr(str));` `    ``}` `}`   `// This code is contributed by Sumit Ghosh`

## Python

 `# Python program`   `import` `sys`   `# A utility function to print a` `# substring str[low..high]` `def` `printSubStr(st, low, high) :` `    ``sys.stdout.write(st[low : high ``+` `1``])` `    ``sys.stdout.flush()` `    ``return` `''`   `# This function prints the longest palindrome` `# substring of st[]. It also returns the length` `# of the longest palindrome` `def` `longestPalSubstr(st) :` `    ``n ``=` `len``(st) ``# get length of input string`   `    ``# table[i][j] will be false if substring ` `    ``# str[i..j] is not palindrome. Else ` `    ``# table[i][j] will be true` `    ``table ``=` `[[``0` `for` `x ``in` `range``(n)] ``for` `y` `                          ``in` `range``(n)] ` `    `  `    ``# All substrings of length 1 are` `    ``# palindromes` `    ``maxLength ``=` `1` `    ``i ``=` `0` `    ``while` `(i < n) :` `        ``table[i][i] ``=` `True` `        ``i ``=` `i ``+` `1` `    `  `    ``# check for sub-string of length 2.` `    ``start ``=` `0` `    ``i ``=` `0` `    ``while` `i < n ``-` `1` `:` `        ``if` `(st[i] ``=``=` `st[i ``+` `1``]) :` `            ``table[i][i ``+` `1``] ``=` `True` `            ``start ``=` `i` `            ``maxLength ``=` `2` `        ``i ``=` `i ``+` `1` `    `  `    ``# Check for lengths greater than 2. ` `    ``# k is length of substring` `    ``k ``=` `3` `    ``while` `k <``=` `n :` `        ``# Fix the starting index` `        ``i ``=` `0` `        ``while` `i < (n ``-` `k ``+` `1``) :` `            `  `            ``# Get the ending index of ` `            ``# substring from starting ` `            ``# index i and length k` `            ``j ``=` `i ``+` `k ``-` `1` `    `  `            ``# checking for sub-string from` `            ``# ith index to jth index iff ` `            ``# st[i + 1] to st[(j-1)] is a ` `            ``# palindrome` `            ``if` `(table[i ``+` `1``][j ``-` `1``] ``and` `                      ``st[i] ``=``=` `st[j]) :` `                ``table[i][j] ``=` `True` `    `  `                ``if` `(k > maxLength) :` `                    ``start ``=` `i` `                    ``maxLength ``=` `k` `            ``i ``=` `i ``+` `1` `        ``k ``=` `k ``+` `1` `    ``print` `"Longest palindrome substring is: "``, printSubStr(st, start,` `                                               ``start ``+` `maxLength ``-` `1``)`   `    ``return` `maxLength ``# return length of LPS`     `# Driver program to test above functions` `st ``=` `"forgeeksskeegfor"` `l ``=` `longestPalSubstr(st)` `print` `"Length is:"``, l`   `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# Solution` `using` `System;`   `class` `GFG {`   `    ``// A utility function to print a` `    ``// substring str[low...( high - (low+1))]` `    ``static` `void` `printSubStr(``string` `str, ``int` `low,` `                            ``int` `high)` `    ``{` `        ``Console.WriteLine(str.Substring(low,` `                                        ``high - low + 1));` `    ``}`   `    ``// This function prints the longest` `    ``// palindrome substring of str[].` `    ``// It also returns the length of the` `    ``// longest palindrome` `    ``static` `int` `longestPalSubstr(``string` `str)` `    ``{`   `        ``// Get length of input string` `        ``int` `n = str.Length;`   `        ``// Table[i, j] will be false if substring` `        ``// str[i..j] is not palindrome. Else` `        ``// table[i, j] will be true` `        ``bool``[, ] table = ``new` `bool``[n, n];`   `        ``// All substrings of length 1 are palindromes` `        ``int` `maxLength = 1;` `        ``for` `(``int` `i = 0; i < n; ++i)` `            ``table[i, i] = ``true``;`   `        ``// Check for sub-string of length 2.` `        ``int` `start = 0;`   `        ``for` `(``int` `i = 0; i < n - 1; ++i) {` `            ``if` `(str[i] == str[i + 1]) {` `                ``table[i, i + 1] = ``true``;` `                ``start = i;` `                ``maxLength = 2;` `            ``}` `        ``}`   `        ``// Check for lengths greater than 2.` `        ``// k is length of substring` `        ``for` `(``int` `k = 3; k <= n; ++k) {`   `            ``// Fix the starting index` `            ``for` `(``int` `i = 0; i < n - k + 1; ++i) {`   `                ``// Get the ending index of substring from` `                ``// starting index i and length k` `                ``int` `j = i + k - 1;`   `                ``// Checking for sub-string from ith index` `                ``// to jth index iff str.charAt(i+1) to` `                ``// str.charAt(j-1) is a palindrome` `                ``if` `(table[i + 1, j - 1] && str[i] == str[j]) {` `                    ``table[i, j] = ``true``;` `                    ``if` `(k > maxLength) {` `                        ``start = i;` `                        ``maxLength = k;` `                    ``}` `                ``}` `            ``}` `        ``}` `        ``Console.Write(``"Longest palindrome substring is: "``);` `        ``printSubStr(str, start, start + maxLength - 1);`   `        ``// Return length of LPS` `        ``return` `maxLength;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(``string``[] args)` `    ``{` `        ``string` `str = ``"forgeeksskeegfor"``;`   `        ``Console.WriteLine(``"Length is: "` `+ longestPalSubstr(str));` `    ``}` `}`   `// This code is contributed by SoumikMondal`

## Javascript

 ``

Output

```Longest palindrome substring is: geeksskeeg
Length is: 10```

Complexity Analysis:

• Time complexity: O(n^2).
Two nested traversals are needed.
• Auxiliary Space: O(n^2).
Matrix of size n*n is needed to store the dp array.

A better space complexity approach can be found in Set-2
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

We can even optimize this code by using map and our time complexity will be O(n^2) and space complexity will be O(n) due to map .

Method 2: Unordered_map.
Approach: The space complexity can be reduced to O(n).

• Idea is that we will traverse every character and store the farthest index of that particular character in map.
• Map does not allow duplicates so if a character comes again then its value in map get updated with latest index.
• Now we will again run a loop to traverse each character in string and pass its index and its map value i.e pass (s[i],map[s[i]]) to find that whether this range is palindrome or not.
• Since we have implemented our code in such a way that it will store maximum length substring.

## C++14

 `#include ` `using` `namespace` `std;`   `void` `longestPalin(string str, ``int` `start, ``int` `end, ``int``& mx,` `                  ``string& ans)` `{` `    ``int` `flag = 1, st = start, en = end;` `    ``while` `(start <= end) {` `        ``if` `(str[start++] != str[end--])` `            ``flag = 0;` `    ``}` `    ``if` `(flag == 1) {` `        ``string res = str.substr(st, en - st + 1);` `        ``if` `(mx < res.size()) {` `            ``ans = res;` `            ``mx = res.size();` `        ``}` `    ``}` `}`   `int` `main()` `{` `    ``string str=``"forgeeksskeegfor"``;` `    ``unordered_map<``char``, ``int``> mp;` `    ``int` `mx = 1;` `    ``string ans=``"h"``;` `    ``for` `(``int` `i = 0; i < str.size(); i++) {` `        ``mp[str[i]] = i;` `    ``}` `    ``for` `(``int` `i = 0; i < str.size(); i++) {` `        ``longestPalin(str, i, mp[str[i]], mx, ans);` `    ``}` `      ``reverse(str.begin(), str.end());` `      ``for` `(``int` `i = 0; i < str.size(); i++) {` `        ``longestPalin(str, i, mp[str[i]], mx, ans);` `      ``}` `    ``cout << ans;` `   ``cout << ``"\nLength is: "` `         ``<< mx;` `}`

Output

```geeksskeeg
Length is: 10```

Complexity Analysis:

Time complexity: O(n^2).
Two nested traversals are needed.
Auxiliary Space: O(n).
Because we are using map (unordered map).

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