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Introduction to Levenshtein distance

Last Updated : 31 Jan, 2024
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Levenshtein distance is a measure of the similarity between two strings, which takes into account the number of insertion, deletion and substitution operations needed to transform one string into the other. 

Operations in Levenshtein distance are:

  • Insertion: Adding a character to string A.
  • Deletion: Removing a character from string A.
  • Replacement: Replacing a character in string A with another character.

Let’s see an example that there is String A: “kitten” which need to be converted in String B: “sitting” so we need to determine the minimum operation required

  • kitten → sitten (substitution of “s” for “k”)
  • sitten → sittin (substitution of “i” for ????”)
  • sittin → sitting (insertion of “g” at the end).

In this case it took three operation do this, so the levenshtein distance will be 3.

  • Upper and lower bounds: If and only if the two strings are identical, the Levenshtein distance is always non-negative and zero. Because it requires completely changing one string into the other through deletions or insertions, the most feasible Levenshtein distance between two strings of length m and n is max(m, n).

Applications of Levenshtein distance:

The Levenshtein distance has various applications in various fields such as:

  • Autocorrect Algorithms: Text editors and messaging applications use the Levenshtein distance in their autocorrect features such as gboard, swift keyboard, etc.
  • Data cleaning: It is widely used in the process of data cleaning and normalization task to reduce redundancy and identify similar records in the data mining process.
  • Data clustering and classification: To identify similar records and cluster them is clustering while identifying similar records and providing them with class labels is classification

Relationship with other edit distance metrics:

Let’s see how Levenshtein distance is different from other distance metrics

  • Damerau-Levenshtein distance: It is similar to the Levenshtein distance, but it just also allows transpositions as an additional operation making it 4 operations.
  • Hamming distance: It can only be applied to strings of equal length, it is used measures the number of positions at which the corresponding characters are different.

Now let’s see its implementation using different approaches in different approaches:

1) Levenshtein distance using a recursive approach

To calculate the Levenshtein distance, In the recursive technique, we will use a simple recursive function. It checks each character in the two strings and performs recursive insertions, removals, and replacements.

Below is the implementation for the above idea:

C++




// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
 
int levenshteinRecursive(const string& str1,
                        const string& str2, int m, int n)
{
 
    // str1 is empty
    if (m == 0) {
        return n;
    }
    // str2 is empty
    if (n == 0) {
        return m;
    }
 
    if (str1[m - 1] == str2[n - 1]) {
        return levenshteinRecursive(str1, str2, m - 1,
                                    n - 1);
    }
 
    return 1
        + min(
 
            // Insert
            levenshteinRecursive(str1, str2, m, n - 1),
            min(
 
                // Remove
                levenshteinRecursive(str1, str2, m - 1,
                                        n),
 
                // Replace
                levenshteinRecursive(str1, str2, m - 1,
                                        n - 1)));
}
 
// Drivers code
int main()
{
    string str1 = "kitten";
    string str2 = "sitting";
 
    // Function Call
    int distance = levenshteinRecursive(
        str1, str2, str1.length(), str2.length());
    cout << "Levenshtein Distance: " << distance << endl;
    return 0;
}


Java




import java.io.*;
 
public class Solution {
 
    public static int levenshteinRecursive(String str1,
                                           String str2, int m, int n) {
        // str1 is empty
        if (m == 0) {
            return n;
        }
          
        // str2 is empty
        if (n == 0) {
            return m;
        }
        if (str1.charAt(m - 1) == str2.charAt(n - 1)) {
            return levenshteinRecursive(str1, str2, m - 1, n - 1);
        }
        return 1 + Math.min(
            // Insert
            levenshteinRecursive(str1, str2, m, n - 1),
            Math.min(
                // Remove
                levenshteinRecursive(str1, str2, m - 1, n),
                 
                // Replace
                levenshteinRecursive(str1, str2, m - 1, n - 1)
            )
        );
    }
 
    public static void main(String[] args) {
        String str1 = "kitten";
        String str2 = "sitting";
 
        int distance = levenshteinRecursive(str1, str2, str1.length(), str2.length());
        System.out.println("Levenshtein Distance: " + distance);
    }
}


Python3




def levenshteinRecursive(str1, str2, m, n):
      # str1 is empty
    if m == 0:
        return n
    # str2 is empty
    if n == 0:
        return m
    if str1[m - 1] == str2[n - 1]:
        return levenshteinRecursive(str1, str2, m - 1, n - 1)
    return 1 + min(
          # Insert    
        levenshteinRecursive(str1, str2, m, n - 1),
        min(
              # Remove
            levenshteinRecursive(str1, str2, m - 1, n),
          # Replace
            levenshteinRecursive(str1, str2, m - 1, n - 1))
    )
 
# Drivers code
str1 = "kitten"
str2 = "sitting"
distance = levenshteinRecursive(str1, str2, len(str1), len(str2))
print("Levenshtein Distance:", distance)


C#




using System;
 
class Program
{
    // Recursive function to calculate Levenshtein Distance
    static int LevenshteinRecursive(string str1, string str2, int m, int n)
    {
        // If str1 is empty, the distance is the length of str2
        if (m == 0)
        {
            return n;
        }
         
        // If str2 is empty, the distance is the length of str1
        if (n == 0)
        {
            return m;
        }
 
        // If the last characters of the strings are the same
        if (str1[m - 1] == str2[n - 1])
        {
            return LevenshteinRecursive(str1, str2, m - 1, n - 1);
        }
 
        // Calculate the minimum of three operations:
        // Insert, Remove, and Replace
        return 1 + Math.Min(
            Math.Min(
                // Insert
                LevenshteinRecursive(str1, str2, m, n - 1),
                // Remove
                LevenshteinRecursive(str1, str2, m - 1, n)
            ),
            // Replace
            LevenshteinRecursive(str1, str2, m - 1, n - 1)
        );
    }
 
    static void Main()
    {
        string str1 = "kitten";
        string str2 = "sitting";
 
        // Function Call
        int distance = LevenshteinRecursive(str1, str2, str1.Length, str2.Length);
        Console.WriteLine("Levenshtein Distance: " + distance);
    }
}


Javascript




// JavaScript code for the above approach:
function levenshteinRecursive(str1, str2, m, n) {
    // Base case: str1 is empty
    if (m === 0) {
        return n;
    }
     
    // Base case: str2 is empty
    if (n === 0) {
        return m;
    }
     
    // If the last characters of both
    // strings are the same
    if (str1[m - 1] === str2[n - 1]) {
        return levenshteinRecursive(str1, str2, m - 1, n - 1);
    }
     
    // Calculate the minimum of three possible
    // operations (insert, remove, replace)
    return 1 + Math.min(
        // Insert
        levenshteinRecursive(str1, str2, m, n - 1),
        // Remove
        levenshteinRecursive(str1, str2, m - 1, n),
        // Replace
        levenshteinRecursive(str1, str2, m - 1, n - 1)
    );
}
 
// Driver code
const str1 = "kitten";
const str2 = "sitting";
 
// Function Call
const distance = levenshteinRecursive(str1, str2, str1.length, str2.length);
console.log("Levenshtein Distance: " + distance);


Output

Levenshtein Distance: 3



Time complexity: O(3^(m+n))
Auxiliary complexity: O(m+n)

2) Levenshtein distance using Iterative with the full matrix approach

The iterative technique with a full matrix uses a 2D matrix to hold the intermediate results of the Levenshtein distance calculation. It begins with empty strings and iteratively fills the matrix row by row. It computes the minimum cost of insertions, deletions, and replacements based on the characters of both strings.

Below is the implementation for the above idea:

C++




// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
 
int levenshteinFullMatrix(const string& str1,
                        const string& str2)
{
    int m = str1.length();
    int n = str2.length();
 
    vector<vector<int> > dp(m + 1, vector<int>(n + 1, 0));
 
    for (int i = 0; i <= m; i++) {
        dp[i][0] = i;
    }
 
    for (int j = 0; j <= n; j++) {
        dp[0][j] = j;
    }
 
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (str1[i - 1] == str2[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1];
            }
            else {
                dp[i][j] = 1
                        + min(
 
                            // Insert
                            dp[i][j - 1],
                            min(
 
                                // Remove
                                dp[i - 1][j],
 
                                // Replace
                                dp[i - 1][j - 1]));
            }
        }
    }
 
    return dp[m][n];
}
 
// Drivers code
int main()
{
    string str1 = "kitten";
    string str2 = "sitting";
 
    // Function Call
    int distance = levenshteinFullMatrix(str1, str2);
    cout << "Levenshtein Distance: " << distance << endl;
    return 0;
}


Java




import java.util.Arrays;
 
public class LevenshteinDistance {
    // Function to calculate Levenshtein Distance between two strings
    public static int levenshteinFullMatrix(String str1, String str2) {
        int m = str1.length();
        int n = str2.length();
 
        // Create a 2D array to store the dynamic programming results
        int[][] dp = new int[m + 1][n + 1];
 
        // Initialize the base cases
        for (int i = 0; i <= m; i++) {
            dp[i][0] = i;
        }
 
        for (int j = 0; j <= n; j++) {
            dp[0][j] = j;
        }
 
        // Fill in the DP array using the recurrence relation
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
                    // Characters match, no operation needed
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    // Characters don't match, consider the minimum of insert, remove, and replace
                    dp[i][j] = 1 + Math.min(
                            // Insert
                            dp[i][j - 1],
                            Math.min(
                                    // Remove
                                    dp[i - 1][j],
                                    // Replace
                                    dp[i - 1][j - 1]));
                }
            }
        }
 
        // Result is stored in the bottom-right cell of the DP array
        return dp[m][n];
    }
 
    public static void main(String[] args) {
        String str1 = "kitten";
        String str2 = "sitting";
 
        // Function Call
        int distance = levenshteinFullMatrix(str1, str2);
         
        // Print the result
        System.out.println("Levenshtein Distance: " + distance);
    }
}


Python3




def levenshteinFullMatrix(str1, str2):
    m = len(str1)
    n = len(str2)
 
    # Initialize a matrix to store the edit distances
    dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]
 
    # Initialize the first row and column with values from 0 to m and 0 to n respectively
    for i in range(m + 1):
        dp[i][0] = i
    for j in range(n + 1):
        dp[0][j] = j
 
    # Fill the matrix using dynamic programming to compute edit distances
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if str1[i - 1] == str2[j - 1]:
                # Characters match, no operation needed
                dp[i][j] = dp[i - 1][j - 1]
            else:
                # Characters don't match, choose minimum cost among insertion, deletion, or substitution
                dp[i][j] = 1 + min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1])
 
    # Return the edit distance between the strings
    return dp[m][n]
 
# Driver code
str1 = "kitten"
str2 = "sitting"
 
# Function Call
distance = levenshteinFullMatrix(str1, str2)
print(f"Levenshtein Distance: {distance}")


C#




using System;
 
class LevenshteinDistance
{
    // Function to calculate Levenshtein Distance using a full matrix approach
    static int LevenshteinFullMatrix(string str1, string str2)
    {
        int m = str1.Length;
        int n = str2.Length;
 
        // Create a matrix to store distances
        int[,] dp = new int[m + 1, n + 1];
 
        // Initialize the first row and column of the matrix
        for (int i = 0; i <= m; i++)
        {
            dp[i, 0] = i; // Number of insertions required for str1 to become an empty string
        }
        for (int j = 0; j <= n; j++)
        {
            dp[0, j] = j; // Number of insertions required for an empty string to become str2
        }
 
        // Fill in the matrix with minimum edit distances
        for (int i = 1; i <= m; i++)
        {
            for (int j = 1; j <= n; j++)
            {
                if (str1[i - 1] == str2[j - 1])
                {
                    dp[i, j] = dp[i - 1, j - 1]; // Characters match, no operation needed
                }
                else
                {
                    // Choose the minimum of insert, delete, or replace operations
                    dp[i, j] = 1 + Math.Min(
                        dp[i, j - 1], // Insertion
                        Math.Min(
                            dp[i - 1, j], // Deletion
                            dp[i - 1, j - 1] // Replacement
                        )
                    );
                }
            }
        }
        return dp[m, n]; // Return the final edit distance
    }
 
    static void Main()
    {
        string str1 = "kitten";
        string str2 = "sitting";
 
        // Calculate Levenshtein Distance between str1 and str2
        int distance = LevenshteinFullMatrix(str1, str2);
        Console.WriteLine("Levenshtein Distance: " + distance);
    }
}


Javascript




// JavaScript code for the above approach:
function levenshteinFullMatrix(str1, str2) {
    const m = str1.length;
    const n = str2.length;
 
    const dp = new Array(m + 1).fill(null).map(() => new Array(n + 1).fill(0));
 
    // Initialize the first row
    // and column of the matrix
    for (let i = 0; i <= m; i++) {
        dp[i][0] = i;
    }
    for (let j = 0; j <= n; j++) {
        dp[0][j] = j;
    }
     
    for (let i = 1; i <= m; i++) {
        for (let j = 1; j <= n; j++) {
            if (str1[i - 1] === str2[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1];
            } else {
                dp[i][j] = 1 + Math.min(
                    // Insert
                    dp[i][j - 1],
                    Math.min(
                        // Remove
                        dp[i - 1][j],
                        // Replace
                        dp[i - 1][j - 1]
                    )
                );
            }
        }
    }
    return dp[m][n];
}
 
// Driver code
const str1 = "kitten";
const str2 = "sitting";
 
// Function Call
const distance = levenshteinFullMatrix(str1, str2);
console.log("Levenshtein Distance:", distance);


Output

Levenshtein Distance: 3



Time complexity: O(m*n)
Auxiliary complexity: O(m*n)

3) Levenshtein distance using Iterative with two matrix rows approach

By simply storing two rows of the matrix at a time, the iterative technique with two matrix rows reduces space complexity. It iterates through the strings row by row, storing the current and past calculations in two rows.

Below is the implementation for the above approach:

C++




// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
 
int levenshteinTwoMatrixRows(const string& str1,
                            const string& str2)
{
    int m = str1.length();
    int n = str2.length();
 
    vector<int> prevRow(n + 1, 0);
    vector<int> currRow(n + 1, 0);
 
    for (int j = 0; j <= n; j++) {
        prevRow[j] = j;
    }
 
    for (int i = 1; i <= m; i++) {
        currRow[0] = i;
 
        for (int j = 1; j <= n; j++) {
            if (str1[i - 1] == str2[j - 1]) {
                currRow[j] = prevRow[j - 1];
            }
            else {
                currRow[j] = 1
                            + min(
 
                                // Insert
                                currRow[j - 1],
                                min(
 
                                    // Remove
                                    prevRow[j],
 
                                    // Replace
                                    prevRow[j - 1]));
            }
        }
 
        prevRow = currRow;
    }
 
    return currRow[n];
}
 
// Drivers code
int main()
{
    string str1 = "kitten";
    string str2 = "sitting";
 
    // Function Call
    int distance = levenshteinTwoMatrixRows(str1, str2);
    cout << "Levenshtein Distance: " << distance;
    return 0;
}


Java




import java.util.Arrays;
 
public class LevenshteinDistance {
    // Method to calculate Levenshtein distance using two matrix rows
    public static int levenshteinTwoMatrixRows(String str1, String str2) {
        int m = str1.length();
        int n = str2.length();
 
        // Initializing two arrays to store the current and previous row values
        int[] prevRow = new int[n + 1];
        int[] currRow = new int[n + 1];
 
        // Initializing the first row with increasing integers
        for (int j = 0; j <= n; j++) {
            prevRow[j] = j;
        }
 
        // Looping through each character of str1
        for (int i = 1; i <= m; i++) {
            // Initializing the first element of the current row with the row number
            currRow[0] = i;
 
            // Looping through each character of str2
            for (int j = 1; j <= n; j++) {
                // If characters are equal, no operation needed, take the diagonal value
                if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
                    currRow[j] = prevRow[j - 1];
                } else {
                    // If characters are not equal, find the minimum value of insert, delete, or replace
                    currRow[j] = 1 + Math.min(currRow[j - 1], Math.min(prevRow[j], prevRow[j - 1]));
                }
            }
 
            // Update prevRow with currRow values
            prevRow = Arrays.copyOf(currRow, currRow.length);
        }
 
        // Return the final Levenshtein distance stored at the bottom-right corner of the matrix
        return currRow[n];
    }
 
    // Main method for testing
    public static void main(String[] args) {
        String str1 = "kitten";
        String str2 = "sitting";
 
        // Function Call
        int distance = levenshteinTwoMatrixRows(str1, str2);
        System.out.println("Levenshtein Distance: " + distance);
    }
}


Python3




# Python program for the above approach
def levenshtein_two_matrix_rows(str1, str2):
    # Get the lengths of the input strings
    m = len(str1)
    n = len(str2)
 
    # Initialize two rows for dynamic programming
    prev_row = [j for j in range(n + 1)]
    curr_row = [0] * (n + 1)
 
    # Dynamic programming to fill the matrix
    for i in range(1, m + 1):
        # Initialize the first element of the current row
        curr_row[0] = i
 
        for j in range(1, n + 1):
            if str1[i - 1] == str2[j - 1]:
                # Characters match, no operation needed
                curr_row[j] = prev_row[j - 1]
            else:
                # Choose the minimum cost operation
                curr_row[j] = 1 + min(
                    curr_row[j - 1],  # Insert
                    prev_row[j],      # Remove
                    prev_row[j - 1]    # Replace
                )
 
        # Update the previous row with the current row
        prev_row = curr_row.copy()
 
    # The final element in the last row contains the Levenshtein distance
    return curr_row[n]
 
# Driver code
if __name__ == "__main__":
    # Example input strings
    str1 = "kitten"
    str2 = "sitting"
 
    # Function call to calculate Levenshtein distance
    distance = levenshtein_two_matrix_rows(str1, str2)
 
    # Print the result
    print("Levenshtein Distance:", distance)
 
# This code is contributed by Susobhan Akhuli


C#




// C# program for the above approach
using System;
 
class LevenshteinDistance {
    // Function to calculate Levenshtein distance between
    // two strings
    static int LevenshteinTwoMatrixRows(string str1,
                                        string str2)
    {
        int m = str1.Length;
        int n = str2.Length;
 
        // Initialize two rows for dynamic programming
        int[] prevRow = new int[n + 1];
        int[] currRow = new int[n + 1];
 
        // Initialization of the first row
        for (int j = 0; j <= n; j++) {
            prevRow[j] = j;
        }
 
        // Dynamic programming to calculate Levenshtein
        // distance
        for (int i = 1; i <= m; i++) {
            // Initialize the current row with the value of
            // i
            currRow[0] = i;
 
            for (int j = 1; j <= n; j++) {
                // If characters are the same, no operation
                // is needed
                if (str1[i - 1] == str2[j - 1]) {
                    currRow[j] = prevRow[j - 1];
                }
                else {
                    // Choose the minimum of three
                    // operations: insert, remove, or
                    // replace
                    currRow[j] = 1
                                 + Math.Min(
                                     // Insert
                                     currRow[j - 1],
                                     Math.Min(
                                         // Remove
                                         prevRow[j],
 
                                         // Replace
                                         prevRow[j - 1]));
                }
            }
 
            // Update the previous row for the next
            // iteration
            Array.Copy(currRow, prevRow, n + 1);
        }
 
        // The bottom-right cell contains the Levenshtein
        // distance
        return currRow[n];
    }
 
    // Main method to test the Levenshtein distance
    // calculation
    static void Main()
    {
        string str1 = "kitten";
        string str2 = "sitting";
 
        // Function Call
        int distance = LevenshteinTwoMatrixRows(str1, str2);
        Console.WriteLine("Levenshtein Distance: "
                          + distance);
    }
}
 
// This code is contributed by Susobhan Akhuli


Javascript




// Function to calculate Levenshtein distance using two matrix rows
function levenshteinTwoMatrixRows(str1, str2) {
    const m = str1.length;
    const n = str2.length;
 
    // Initialize two arrays to represent the matrix rows
    let prevRow = new Array(n + 1).fill(0);
    let currRow = new Array(n + 1).fill(0);
 
    // Initialize the first row with consecutive numbers
    for (let j = 0; j <= n; j++) {
        prevRow[j] = j;
    }
 
    // Dynamic programming to fill the matrix
    for (let i = 1; i <= m; i++) {
        currRow[0] = i;
 
        for (let j = 1; j <= n; j++) {
            // Check if characters at the current positions are equal
            if (str1[i - 1] === str2[j - 1]) {
                currRow[j] = prevRow[j - 1]; // No operation required
            } else {
                // Choose the minimum of three possible operations (insert, remove, replace)
                currRow[j] = 1 + Math.min(
                    currRow[j - 1],   // Insert
                    prevRow[j],       // Remove
                    prevRow[j - 1]    // Replace
                );
            }
        }
 
        // Update the previous row with the current row for the next iteration
        prevRow = [...currRow];
    }
 
    // The result is the value at the bottom-right corner of the matrix
    return currRow[n];
}
 
// Driver code
const str1 = "kitten";
const str2 = "sitting";
 
// Function call to calculate Levenshtein distance
const distance = levenshteinTwoMatrixRows(str1, str2);
 
// Print the result
console.log("Levenshtein Distance:", distance);


Output

Levenshtein Distance: 3


Time complexity: O(m*n)
Auxiliary Space: O(n)



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