Longest Palindromic Substring in Dynamic Programming

The "Longest Palindromic Substring" problem can also be solved using Dynamic Programming, which is a method to efficiently solve a complex problem by breaking it down into simpler subproblems and storing the solutions of these subproblems to avoid redundant computations. This approach is particularly useful in this problem to identify palindromic substrings and build upon them. Let's explore how dynamic programming is applied here.

Dynamic Programming Approach:


  1. Subproblem: The longest palindromic substring can be found by solving smaller instances of the same problem, i.e., finding palindromic substrings within the given string.
  2. State: A 2D boolean array dp[i][j] is used, where dp[i][j] is true if the substring from index i to j in the given string is a palindrome.


  1. Initialization: Start by creating a 2D array dp with dimensions equal to the length of the string, initialized to false.
  2. Base Cases:
    • Every single character is a palindrome by itself, so dp[i][i] is true for all i.
    • For two-character substrings, dp[i][i+1] is true if the two characters are the same.
  3. Filling the Table: Fill the dp table in a bottom-up manner. The substring s[i:j] is a palindrome if s[i] == s[j] and s[i+1:j-1] is also a palindrome (i.e., dp[i+1][j-1] is true).
  4. Result: Keep track of the longest palindromic substring's length and starting index as the table is filled.

Time and Space Complexity:

  • Time Complexity: O(n²), since we fill a table of size n².
  • Space Complexity: O(n²), for the 2D array.

Example Implementation in Python:

def longest_palindromic_substring(s):
    n = len(s)
    if n == 0:
        return ""

    # Initialize DP table
    dp = [[False for _ in range(n)] for _ in range(n)]
    start = 0
    max_length = 1

    # Every single character is a palindrome
    for i in range(n):
        dp[i][i] = True

    # Check for two-character palindromes
    for i in range(n - 1):
        if s[i] == s[i + 1]:
            dp[i][i + 1] = True
            start = i
            max_length = 2

    # Check for palindromes longer than 2
    for length in range(3, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            if s[i] == s[j] and dp[i + 1][j - 1]:
                dp[i][j] = True
                start = i
                max_length = length

    return s[start:start + max_length]

In this implementation:

  • We systematically build the solution for smaller substrings and use these solutions to solve larger substrings.
  • The dp table is filled in a way that ensures that all smaller substrings are processed before the larger ones.
  • The final result is the longest palindromic substring found during the table filling process.