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

Edit Distance Problem Using Dynamic Programming

Strings

Contents

ProblemExamplesVisualizationSolutionAlgorithmCode

Problem Statement❯

The Edit Distance problem asks: Given two strings word1 and word2, what is the minimum number of operations required to convert word1 into word2?

  • Allowed operations are:
    • Insert a character
    • Delete a character
    • Replace a character
  • Each operation counts as 1 step.

The goal is to find the minimum number of such operations needed to make the two strings equal.

Examples❯

Word 1 Word 2 Edit Distance Description
"horse" "ros" 3 Replace 'h' → 'r', remove 'r', remove 'e'
"intention" "execution" 5 Requires multiple inserts, deletes, and replacements
"abc" "abc" 0 Both strings are already equal
"abc" "def" 3 All characters need to be replaced
"abcd" "abc" 1 Only one deletion is needed
"abc" "abcd" 1 Only one insertion is needed
"" "abc" 3 Empty word1, need 3 insertions
"abc" "" 3 Empty word2, need 3 deletions
"" "" 0 Both strings are empty, no operations needed

Visualization Player

Solution❯

To solve the Edit Distance problem, we need to think about how one string can be gradually transformed into another. We are allowed three operations: insert, delete, or replace a character. Each of these costs 1 unit, and the goal is to reach the target string in the fewest steps.

Let's say we are trying to convert word1 to word2:

  • If both strings are empty, then no operations are needed — the distance is 0.
  • If one string is empty and the other is not, then the only option is to insert or delete characters one-by-one. For example, converting "" to "abc" takes 3 insertions.
  • If the characters at the current position in both strings are the same, we don’t need to do anything — we just move to the next characters.
  • But if the characters are different, we have three choices:
    • Insert: Imagine we insert the next character from word2 into word1.
    • Delete: Imagine we delete the current character from word1.
    • Replace: Imagine we replace the current character of word1 with the character from word2.
    We pick the operation with the smallest cost and add 1 to it.

To make this efficient, we store the results of already-solved subproblems in a 2D table. Each cell in the table dp[i][j] tells us the edit distance between the first i characters of word1 and the first j characters of word2.

By building this table from the base cases (like when one of the strings is empty), we ensure we don’t compute the same values again and again. The final answer will be at dp[m][n], where m and n are the lengths of word1 and word2 respectively.

This approach ensures we consider all possibilities and always take the minimum number of operations needed, making it suitable even for large strings.

Algorithm Steps❯

  1. Let word1 and word2 be the input strings, with lengths m and n respectively.
  2. Create a 2D table dp of dimensions (m+1) x (n+1), where dp[i][j] represents the edit distance between the first i characters of word1 and the first j characters of word2.
  3. Initialize the first column: for each i from 0 to m, set dp[i][0] = i (representing i deletions).
  4. Initialize the first row: for each j from 0 to n, set dp[0][j] = j (representing j insertions).
  5. For each i from 1 to m and for each j from 1 to n:
    1. If word1[i-1] equals word2[j-1], then set dp[i][j] = dp[i-1][j-1].
    2. Otherwise, set dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]).
  6. The edit distance is dp[m][n].

Code

Python
Java
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
def edit_distance(word1, word2):
    m, n = len(word1), len(word2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    for i in range(m + 1):
        dp[i][0] = i
    for j in range(n + 1):
        dp[0][j] = j
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if word1[i - 1] == word2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            else:
                dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1])
    return dp[m][n]

if __name__ == '__main__':
    print("Edit Distance between 'kitten' and 'sitting':", edit_distance("kitten", "sitting"))
    print("Edit Distance between 'horse' and 'ros':", edit_distance("horse", "ros"))