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Java/src/main/java/com/thealgorithms/dynamicprogramming/LevenshteinDistance.java
SOZEL dda3c9cb59 Refactor Levenshtein distance implementation (#5138)
* ref: refactor Levenshtein distance implementation
- Rewrite the original levenshtein distance implementation in functional style
- Add optimized version of levenshtein distance

* ref: make `LevenshteinDistance` class a proper utility

* ref: remove duplicated test data

* ref: update tests

---

Co-authored-by: Piotr Idzik <65706193+vil02@users.noreply.github.com>
2024-05-04 11:13:30 +02:00

85 lines
4.0 KiB
Java

package com.thealgorithms.dynamicprogramming;
import java.util.stream.IntStream;
/**
* Provides functions to calculate the Levenshtein distance between two strings.
*
* The Levenshtein distance is a measure of the similarity between two strings by calculating the minimum number of single-character
* edits (insertions, deletions, or substitutions) required to change one string into the other.
*/
public final class LevenshteinDistance {
private LevenshteinDistance() {
}
/**
* Calculates the Levenshtein distance between two strings using a naive dynamic programming approach.
*
* This function computes the Levenshtein distance by constructing a dynamic programming matrix and iteratively filling it in.
* It follows the standard top-to-bottom, left-to-right approach for filling in the matrix.
*
* @param string1 The first string.
* @param string2 The second string.
* @return The Levenshtein distance between the two input strings.
*
* Time complexity: O(nm),
* Space complexity: O(nm),
*
* where n and m are lengths of `string1` and `string2`.
*
* Note that this implementation uses a straightforward dynamic programming approach without any space optimization.
* It may consume more memory for larger input strings compared to the optimized version.
*/
public static int naiveLevenshteinDistance(final String string1, final String string2) {
int[][] distanceMatrix = IntStream.rangeClosed(0, string1.length()).mapToObj(i -> IntStream.rangeClosed(0, string2.length()).map(j -> (i == 0) ? j : (j == 0) ? i : 0).toArray()).toArray(int[][] ::new);
IntStream.range(1, string1.length() + 1).forEach(i -> IntStream.range(1, string2.length() + 1).forEach(j -> {
final int cost = (string1.charAt(i - 1) == string2.charAt(j - 1)) ? 0 : 1;
distanceMatrix[i][j] = Math.min(distanceMatrix[i - 1][j - 1] + cost, Math.min(distanceMatrix[i][j - 1] + 1, distanceMatrix[i - 1][j] + 1));
}));
return distanceMatrix[string1.length()][string2.length()];
}
/**
* Calculates the Levenshtein distance between two strings using an optimized dynamic programming approach.
*
* This edit distance is defined as 1 point per insertion, substitution, or deletion required to make the strings equal.
*
* @param string1 The first string.
* @param string2 The second string.
* @return The Levenshtein distance between the two input strings.
*
* Time complexity: O(nm),
* Space complexity: O(n),
*
* where n and m are lengths of `string1` and `string2`.
*
* Note that this implementation utilizes an optimized dynamic programming approach, significantly reducing the space complexity from O(nm) to O(n), where n and m are the lengths of `string1` and `string2`.
*
* Additionally, it minimizes space usage by leveraging the shortest string horizontally and the longest string vertically in the computation matrix.
*/
public static int optimizedLevenshteinDistance(final String string1, final String string2) {
if (string1.isEmpty()) {
return string2.length();
}
int[] previousDistance = IntStream.rangeClosed(0, string1.length()).toArray();
for (int j = 1; j <= string2.length(); j++) {
int prevSubstitutionCost = previousDistance[0];
previousDistance[0] = j;
for (int i = 1; i <= string1.length(); i++) {
final int deletionCost = previousDistance[i] + 1;
final int insertionCost = previousDistance[i - 1] + 1;
final int substitutionCost = (string1.charAt(i - 1) == string2.charAt(j - 1)) ? prevSubstitutionCost : prevSubstitutionCost + 1;
prevSubstitutionCost = previousDistance[i];
previousDistance[i] = Math.min(deletionCost, Math.min(insertionCost, substitutionCost));
}
}
return previousDistance[string1.length()];
}
}