Binary TreesBinary Trees36

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  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

Edit Distance Problem
Using Dynamic Programming



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?

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

Examples

Word 1Word 2Edit DistanceDescription
horseros3Replace 'h' → 'r', remove 'r', remove 'e'
intentionexecution5Requires multiple inserts, deletes, and replacements
abcabc0Both strings are already equal
abcdef3All characters need to be replaced
abcdabc1Only one deletion is needed
abcabcd1Only one insertion is needed
""abc3Empty word1, need 3 insertions
abc""3Empty word2, need 3 deletions
""""0Both strings are empty, no operations needed

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.

Visualization

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"))


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