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

Largest Palindromic Substring (Without DP)

Problem Statement

Given a string, your task is to find the longest palindromic substring present in it. A substring is a contiguous sequence of characters within the string, and a palindrome is a string that reads the same forward and backward.

Note: You must solve this problem without using Dynamic Programming (DP). Center expansion technique is allowed.

Examples

Input String Longest Palindromic Substring Description
"abaxyzzyxf" xyzzyx Longest palindrome in the middle
"babad" bab "bab" and "aba" are both valid; return any one
"cbbd" bb Longest even-length palindrome
"a" a Single character string is itself a palindrome
"ac" a No multi-letter palindrome, return any single character
"aaaa" aaaa All characters same, entire string is the palindrome
"abcde" a No repeated characters, return any single character
"" Empty string has no palindromic substring

Solution

To find the longest palindromic substring without using dynamic programming, we use a technique called center expansion. The idea is simple and intuitive: every palindrome is centered around a character (or between two characters), so we expand around each possible center and keep track of the longest one found.

Understanding Through Cases

Case 1: The palindrome has an odd length
Here, we consider each character as the center. For example, in the word racecar, the character 'e' is at the center of the palindrome. We expand both left and right from this center and compare characters to check if it forms a palindrome.

Case 2: The palindrome has an even length
Some palindromes are centered between two characters, like abba. In this case, we expand around the pair of equal characters and compare outward on both sides.

Case 3: Multiple palindromes
If there are multiple palindromic substrings of the same maximum length, we can return any one of them. For example, in babad, both "bab" and "aba" are valid answers.

Case 4: No palindromes longer than 1
If the string contains no repeated characters or palindromes, like abc, then each character is a palindrome of length 1. We can return any one of them.

Case 5: All characters are the same
In cases like aaaa, the entire string is a palindrome. Expanding from any center will eventually cover the whole string.

Case 6: Empty string
If the input string is empty, we return an empty string as there’s no palindromic substring to be found.

Why This Works

By checking for palindromes centered at every character (and between characters), we cover all possibilities. The approach avoids building a 2D table (used in DP), and runs efficiently in O(n²) time and O(1) space, which is optimal for a non-DP solution.

It’s a powerful technique that gives the correct answer without any complex setup—just repeated expansion and comparison.

Algorithm Steps

  1. Initialize two variables start and end to track the longest palindrome range.
  2. For each index i in the string:
  3. → Expand around center with one character (odd-length).
  4. → Expand around center with two characters (even-length).
  5. → Update start and end if a longer palindrome is found.
  6. Return the substring from start to end (inclusive).

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)If the string is already a palindrome, each center will expand quickly and minimal comparisons are made.
Average CaseO(n^2)Each character is treated as a center and expanded outward, leading to quadratic time in general.
Worst CaseO(n^2)In the worst case, each expansion checks almost the entire string (e.g., for input like 'aaaaaaa').

Space Complexity

O(1)

Explanation: No extra space is used except for a few integer variables.