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

Word Ladder - 1 Shortest Transformation Using Graph

Problem Statement

Given two words startWord and targetWord, and a list of unique words wordList, the Word Ladder problem asks you to find the shortest sequence of transformations that converts startWord into targetWord. Each transformation must change exactly one character and result in a valid word in wordList.

If such a sequence exists, return its length. Otherwise, return 0.

Examples

startWord targetWord wordList Output Description
hit cog ["hot","dot","dog","lot","log","cog"] 5 hit → hot → dot → dog → cog
hit cog ["hot","dot","dog","lot","log"] 0 No valid path to targetWord
a c ["a", "b", "c"] 2 a → c
lost cost ["lost","cost"] 2 lost → cost
game thee ["gave","gape","tape","tame","thee"] 0 No valid transformations

Solution

Understanding the Problem

We are given a startWord, a targetWord, and a wordList. The goal is to transform the startWord into the targetWord by changing only one character at a time, with the restriction that each transformed word must be a valid word in the given wordList.

This problem can be visualized as a graph where each word is a node, and an edge exists between two words if they differ by exactly one character. Our task is to find the shortest path (in terms of number of transformations) from the startWord to the targetWord. The most suitable algorithm for this task is Breadth-First Search (BFS), which efficiently finds the shortest path in an unweighted graph.

Step-by-Step Solution with Example

Step 1: Check if transformation is possible

If the targetWord is not in the wordList, then there’s no way to reach it via valid transformations. So, we return 0 immediately.

Step 2: Treat words as graph nodes

Each word becomes a node in a graph. Two words are connected by an edge if they differ by exactly one letter. For example, "hot" and "dot" are neighbors, but "hot" and "dog" are not.

Step 3: Use BFS for shortest path

We initialize a queue for BFS with the pair (startWord, level = 1). The level variable tracks how many transformations we've made so far. We also convert the wordList to a set for O(1) lookup and remove the startWord if it's present.

Step 4: Generate possible transformations

For each word we dequeue, we try changing every character to every letter from 'a' to 'z'. For example, from "hot", we generate words like "aot", "bot", ..., "zot", then "hat", "hbt", ..., "hzt", and finally "hoa", "hob", ..., "hoz".

If a transformed word exists in the wordList set, it means we’ve found a valid transformation. We enqueue this new word with level + 1 and remove it from the set to avoid revisiting.

Step 5: Stop when targetWord is found

If during BFS we dequeue a word that matches the targetWord, we return the current level. This ensures the minimum number of steps due to the nature of BFS (shortest path search).

Step 6: Return 0 if targetWord is unreachable

If the BFS completes and we never encounter the targetWord, that means there is no valid transformation path, and we return 0.

Edge Cases

  • targetWord not in wordList: Return 0 immediately.
  • startWord is same as targetWord: Usually undefined by the problem, but often treated as requiring at least one transformation, so result should be 0 if no path exists.
  • wordList is empty: No transformation is possible, return 0.
  • All words are the same length but differ significantly: The solution should still handle them with correct transformation checks.

Finally

This problem is a great example of using BFS for shortest path discovery in unweighted graphs. By thinking of each word as a node and transformations as edges, we can apply standard graph traversal techniques. Always remember to handle edge cases like missing target words or cycles due to repeated transformations.

With a clear understanding of BFS and word transformation rules, even beginners can confidently approach and solve the Word Ladder problem.

Algorithm Steps

  1. If targetWord is not in wordList, return 0.
  2. Initialize a queue with a pair: (startWord, 1) where 1 represents the initial level.
  3. Convert wordList into a set for O(1) lookups.
  4. While the queue is not empty:
    1. Dequeue the current word and its level.
    2. If the current word is equal to targetWord, return the level.
    3. Generate all possible one-letter transformations.
    4. If the transformed word is in wordList, enqueue it with level+1 and remove it from the set.
  5. If BFS completes without finding targetWord, return 0.

Code

JavaScript
function wordLadder(startWord, targetWord, wordList) {
  const wordSet = new Set(wordList);
  if (!wordSet.has(targetWord)) return 0;

  const queue = [[startWord, 1]];
  while (queue.length > 0) {
    const [word, level] = queue.shift();
    if (word === targetWord) return level;

    for (let i = 0; i < word.length; i++) {
      for (let c = 97; c <= 122; c++) {
        const char = String.fromCharCode(c);
        const newWord = word.slice(0, i) + char + word.slice(i + 1);

        if (wordSet.has(newWord)) {
          queue.push([newWord, level + 1]);
          wordSet.delete(newWord);
        }
      }
    }
  }

  return 0;
}

console.log("Length of shortest transformation:", wordLadder("hit", "cog", ["hot","dot","dog","lot","log","cog"]));

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(m * n)Where n is the number of words and m is the length of each word. In the best case, the target is found early in the BFS traversal, but we still generate m*26 transformations per word.
Average CaseO(m * n)For each word in the wordList, up to m*26 transformations are checked, and each valid transformation is enqueued once.
Worst CaseO(m * n)All words are visited and all possible transformations are generated, each requiring O(m) time. So, total operations are O(m * 26 * n) ≈ O(m * n).

Space Complexity

O(n)

Explanation: We store the wordList as a set of size n, and use a queue which can hold up to n elements in the worst case.


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