Longest Palindromic Substring | Set 1

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

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

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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.

// A C++ solution for longest palindrome 
#include <bits/stdc++.h> 
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; 
} 

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.

Maintain a boolean table[n][n] that is filled in bottom up manner.
The value of table[i][j] is true, if the substring is palindrome, otherwise false.
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.
Otherwise, the value of table[i][j] is made false.
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/C++

// A C++ dynamic programming 
// solution for longest palindrome 
  
#include <bits/stdc++.h> 
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 

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.

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C/C++

Java

Python

C#

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