Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

Longest Palindromic Substring using Dynamic Programming

Problem Statement

Given a string s, your task is to find the longest palindromic substring within it.

  • A palindrome is a sequence of characters that reads the same forward and backward.
  • Your goal is to return the longest continuous segment of the input string that forms a palindrome.

If multiple substrings of the same maximum length exist, you can return any one of them.

If the input string is empty, return an empty string.

Examples

Input String Output Description
"babad" "bab" or "aba" Both "bab" and "aba" are valid palindromes of length 3
"cbbd" "bb" The longest palindrome is "bb" of length 2
"a" "a" Single character is always a palindrome
"ac" "a" or "c" No longer palindrome exists, return any single character
"racecar" "racecar" The entire string is a palindrome
"aacabdkacaa" "aca" "aca" is the longest palindromic substring
"" "" Empty input string returns empty output

Visualization Player

Solution

To find the longest palindromic substring in a given string, we look for sequences that read the same forward and backward. This problem is a perfect candidate for dynamic programming because we can build answers to bigger problems (long substrings) using answers to smaller problems (short substrings).

What’s the core idea?

A substring is a palindrome if the characters at the ends are equal and the part in between is also a palindrome. So we use a table (2D array) to record whether each substring s[i..j] is a palindrome.

How does dynamic programming help?

We create a 2D boolean table dp[i][j] where each cell tells us if the substring from index i to j is a palindrome. Once we know the result of smaller substrings, we can reuse that information to determine if a larger substring is also a palindrome.

How do we fill the table?

  • Substrings of length 1: Every single character is a palindrome. So we set dp[i][i] = true.
  • Substrings of length 2: These need special handling. We can't apply the general DP rule (checking the inner substring) because there's nothing between two characters. So instead, we just check if s[i] == s[i+1], and if so, mark dp[i][i+1] = true.

Why do this separately? If we applied the usual rule dp[i+1][j-1] here, we’d be accessing invalid indices when i+1 > j-1. So we handle 2-letter substrings directly and use them as a base to build longer palindromes like "abba".

  • Substrings of length ≥ 3: For any substring s[i..j], we check:
    • if s[i] == s[j]
    • and if dp[i+1][j-1] is true (i.e., the inner part is already a palindrome)
    If both conditions are met, we mark dp[i][j] = true.

How do we track the result?

While filling the dp table, we keep track of the longest palindrome found so far. If we find a new valid palindrome that’s longer than the previous one, we update the starting index and max length. At the end, we extract and return the longest substring using these values.

What about special cases?

  • Empty input: If the input string is empty, we return an empty string.
  • Multiple answers: For strings like "babad", both "bab" and "aba" are valid. We can return either.

Why this works efficiently

Without DP, we might check the same substring repeatedly. With DP, once we compute whether s[i..j] is a palindrome, we store it and never recompute it again. This gives us an efficient O(n²) solution, which works well for strings up to 1000 characters.

Algorithm Steps

  1. Let n be the length of the input string s.
  2. Create a 2D boolean array dp[n][n].
  3. Every single character is a palindrome, so set dp[i][i] = true for all i.
  4. Check all substrings of length 2 and mark dp[i][i+1] = true if s[i] == s[i+1].
  5. For substrings longer than 2, set dp[i][j] = true if s[i] == s[j] and dp[i+1][j-1] == true.
  6. Track the start index and max length of the longest palindrome found.
  7. Return the substring using start index and max length.

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
public class LongestPalindromicSubstringDP {
  public static String longestPalindrome(String s) {
    int n = s.length();
    if (n <= 1) return s;

    boolean[][] dp = new boolean[n][n];
    int start = 0, maxLength = 1;

    // All substrings of length 1 are palindromes
    for (int i = 0; i < n; i++) dp[i][i] = true;

    // Check for substrings of length 2
    for (int i = 0; i < n - 1; i++) {
      if (s.charAt(i) == s.charAt(i + 1)) {
        dp[i][i + 1] = true;
        start = i;
        maxLength = 2;
      }
    }

    // Check substrings longer than 2
    for (int len = 3; len <= n; len++) {
      for (int i = 0; i <= n - len; i++) {
        int j = i + len - 1;
        if (s.charAt(i) == s.charAt(j) && dp[i + 1][j - 1]) {
          dp[i][j] = true;
          if (len > maxLength) {
            start = i;
            maxLength = len;
          }
        }
      }
    }
    return s.substring(start, start + maxLength);
  }

  public static void main(String[] args) {
    String input = "babad";
    System.out.println("Longest Palindromic Substring: " + longestPalindrome(input));
  }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n^2)We fill an n x n DP table and check each substring for palindromicity.
Average CaseO(n^2)All substrings are checked; DP avoids redundant recalculations.
Worst CaseO(n^2)The algorithm needs to check all possible substrings using nested loops.

Space Complexity

O(n^2)

Explanation: We use a 2D boolean table to store whether substrings are palindromes.