Edit Distance
Given two strings, return the minimum number of operations (insert, delete, replace) to convert one to the other.
Example:
horse ros
3
- The input strings are "horse" and "ros", and we need to find the minimum number of operations to convert "horse" to "ros".
- We start by comparing the two strings: "horse" has 5 characters and "ros" has 3 characters, so we need to delete 2 characters from "horse".
- To convert "horse" to "ros", we can delete "h" and "e", and replace "o" is not needed as it is present in both, but we need to replace the other characters: the resulting operations are 1 replacement (for "r" to match) and 2 deletions (for "h" and "e"), totaling 3 operations.
- The final output is 3, which is the minimum number of operations required to convert "horse" to "ros".
Constraints:
- 0 <= len(word1), len(word2) <= 500
- Strings consist of lowercase English letters
Background Knowledge
The Edit Distance problem is a classic example of a Dynamic Programming problem. Dynamic programming is a method for solving complex problems by breaking them down into simpler subproblems, solving each subproblem only once, and storing the solutions to subproblems to avoid redundant computation. In the context of edit distance, we're dealing with strings, which are sequences of characters. The goal is to find the minimum number of operations (insertions, deletions, and substitutions) required to transform one string into another.
To understand this problem, it's essential to be familiar with the concept of sequence alignment, which is a fundamental problem in computer science and bioinformatics. Sequence alignment involves comparing two or more sequences to identify similarities and differences. In the case of edit distance, we're interested in finding the minimum number of operations required to align two strings. This problem has numerous applications, including data compression, text search, and genomic sequence analysis.
The key concept in solving the edit distance problem is to recognize that the solution to the larger problem depends on the solutions to smaller subproblems. Specifically, the edit distance between two strings can be computed by considering the edit distances between their prefixes. This is where dynamic programming comes into play, as it allows us to store the solutions to these subproblems in a table and reuse them to avoid redundant computation.
Algorithm/Approach
The general approach to solving the edit distance problem involves using a dynamic programming algorithm. The algorithm will typically involve creating a 2D table to store the edit distances between prefixes of the two input strings. The table will be filled in row by row, with each cell representing the edit distance between the corresponding prefixes of the two strings. The final solution will be stored in the bottom-right cell of the table.
Step-by-Step Strategy
To implement the solution, follow these steps:
- Create a 2D table with dimensions (m+1)×(n+1), where m and n are the lengths of the two input strings.
- Initialize the first row and column of the table to represent the edit distances between the empty string and the prefixes of the two input strings.
- Fill in the rest of the table row by row, using the following recurrence relation: d[i][j]=min(d[i−1][j−1]+cost(i,j),d[i−1][j]+1,d[i][j−1]+1), where cost(i,j) is the cost of substituting the ith character of the first string with the jth character of the second string.
- The final solution will be stored in the bottom-right cell of the table, d[m][n].
Common Pitfalls
When implementing the solution, watch out for the following common pitfalls:
- Forgetting to initialize the first row and column of the table correctly.
- Using the wrong recurrence relation to fill in the table.
- Not handling the base cases correctly (e.g., when one or both of the input strings are empty).
Time & Space Complexity
The time complexity of the edit distance algorithm is O(mn), where m and n are the lengths of the two input strings. This is because we need to fill in a 2D table with dimensions (m+1)×(n+1). The space complexity is also O(mn), as we need to store the entire table in memory.