The** Levenshtein distance** also called the **Edit distance**, is the minimum number of operations required to transform one string to another.

Typically, three types of operations are performed (one at a time) :

- Replace a character.
- Delete a character.
- Insert a character.

**Examples:**

Input: str1 = “glomax”, str2 = “folmax”

Output: 3

str1 is converted to str2 by replacing ‘g’ with ‘o’, deleting the second ‘o’, and inserting ‘f’ at the beginning. There is no way to do it with fewer than three edits.

Input: s1 = “GIKY”, s2 = “GEEKY”

Output: 2

s1 is converted to s2 by inserting ‘E’ right after ‘G’,and replacing ‘I’ with ‘E’.

**This problem can be done in two ways :**

- Using Recursion.
- Using Dynamic Programming.

**Method 1: Recursive Approach **

Let’s consider by taking an example

Given two strings s1 = “sunday” and s2 = “saturday”. We want to convert “sunday” into “saturday” with minimum edits.

- Consider ‘i’ and ‘j’ as the upper-limit indices of substrings generated using s1 and s2.
- Let us pick i = 2 and j = 4 i.e. prefix strings are ‘su’ and ‘satu’ respectively (assume the strings indices start at 1). The rightmost characters can be aligned in three different possible ways.
- Possible
**Case 1**: Align the characters ‘u’ and ‘u’. They are equal, no edit is required. We still left with the problem of i = 1 and j = 3, so we should proceed to find Levenshtein distance(i-1, j-1). - Possible
**Case 2**(Deletion): Align the right character from the first string and no character from the second string. We need a deletion here. We still left with problem of i = 1 and j = 4, so we should proceed finding Levenshtein distance(i-1, j). - Possible
**Case 3**(Insertion): Align the right character from the second string and no character from the first string. We need an insertion here. We still left with the problem of i = 2 and j = 3, so we should proceed to find Levenshtein distance(i, j-1). - We assume that the character to be inserted in the first string is the same as the right character of the second string.
- Possible
**Case 4**(Replacement): Align right characters from the first string as well as from the second string. We need a substitution here. We still left with problem of i = 1 and j = 3, so we should proceed finding Levenshtein distance(i-1, j-1). - We assume that the replaced character in the first string is the same as the right character of the second string.
- We have to find the minimum of all the possible three cases.

**Recursive Implementation:**

## Java

`// Java implementation of recursive Levenshtein distance` `// calculation` ` ` `import` `java.util.*;` `class` `LevenshteinDistanceRecursive {` ` ` ` ` `static` `int` `compute_Levenshtein_distance(String str1,` ` ` `String str2)` ` ` `{` ` ` `// If str1 is empty, all` ` ` `// characters of str2 are` ` ` `// inserted into str1, which is` ` ` `// of the only possible method of` ` ` `// conversion with minimum` ` ` `// operations.` ` ` ` ` `if` `(str1.isEmpty())` ` ` `{` ` ` `return` `str2.length();` ` ` `}` ` ` ` ` `// If str2 is empty, all` ` ` `// characters of str1 are` ` ` `// removed, which is the` ` ` `// only possible` ` ` `// method of conversion with minimum` ` ` `// operations.` ` ` ` ` `if` `(str2.isEmpty()) ` ` ` `{` ` ` `return` `str1.length();` ` ` `}` ` ` ` ` `// calculate the number of distinct characters to be` ` ` `// replaced in str1` ` ` `// by recursively traversing each substring` ` ` ` ` `int` `replace = compute_Levenshtein_distance(` ` ` `str1.substring(` `1` `), str2.substring(` `1` `))` ` ` `+ NumOfReplacement(str1.charAt(` `0` `),str2.charAt(` `0` `));` ` ` ` ` `// calculate the number of insertions in str1` ` ` `// recursively` ` ` `int` `insert = compute_Levenshtein_distance(` ` ` `str1, str2.substring(` `1` `))+ ` `1` `;` ` ` ` ` `// calculate the number of deletions in str1` ` ` `// recursively` ` ` `int` `delete = compute_Levenshtein_distance(` ` ` `str1.substring(` `1` `), str2)+ ` `1` `;` ` ` ` ` `// returns minimum of three operatoins` ` ` ` ` `return` `minm_edits(replace, insert, delete);` ` ` `}` ` ` ` ` `static` `int` `NumOfReplacement(` `char` `c1, ` `char` `c2)` ` ` `{` ` ` `// check for distinct characters` ` ` `// in str1 and str2` ` ` ` ` `return` `c1 == c2 ? ` `0` `: ` `1` `;` ` ` `}` ` ` ` ` `static` `int` `minm_edits(` `int` `... nums)` ` ` `{` ` ` `// receives the count of different` ` ` `// operations performed and returns the` ` ` `// minimum value among them.` ` ` ` ` `return` `Arrays.stream(nums).min().orElse(` ` ` `Integer.MAX_VALUE);` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `String s1 = ` `"glomax"` `;` ` ` `String s2 = ` `"folmax"` `;` ` ` ` ` `System.out.println(compute_Levenshtein_distance(s1, s2));` ` ` `}` `}` |

**Output**

3

**Time Complexity: **O(3^n) because at each step, we branch-off into three recursive calls. Here, ‘n’ is the length of the first string.

**Method 2: Dynamic Programming Approach **

If we draw the recursion tree of the above solution, we can see that the same sub-problems are getting computed again and again. We know that Dynamic Programming comes to the picture when subproblem solutions can be memoized rather than computed again and again.

- The
*Memoized*version follows the top-down approach since we first break the problems into subproblems and then calculate and store values. - We can also solve this problem in a bottom-up approach. In a bottom-up manner, we solve the sub-problems first, then solve larger sub-problems from them.

** Dynamic Programming Implementation (Optimised approach)**

## Java

`// Java implementation of Levenshtein distance calculation` `// Using Dynamic Programming (Optimised solution)` ` ` `import` `java.util.*;` `class` `LevenshteinDistanceDP {` ` ` ` ` `static` `int` `compute_Levenshtein_distanceDP(String str1,` ` ` `String str2)` ` ` `{` ` ` ` ` `// A 2-D matrix to store previously calculated` ` ` `// answers of subproblems in order` ` ` `// to obtain the final` ` ` ` ` `int` `[][] dp = ` `new` `int` `[str1.length() + ` `1` `][str2.length() + ` `1` `];` ` ` ` ` `for` `(` `int` `i = ` `0` `; i <= str1.length(); i++) ` ` ` `{` ` ` `for` `(` `int` `j = ` `0` `; j <= str2.length(); j++) {` ` ` ` ` `// If str1 is empty, all characters of` ` ` `// str2 are inserted into str1, which is of` ` ` `// the only possible method of conversion` ` ` `// with minimum operations.` ` ` `if` `(i == ` `0` `) {` ` ` `dp[i][j] = j;` ` ` `}` ` ` ` ` `// If str2 is empty, all characters of str1` ` ` `// are removed, which is the only possible` ` ` `// method of conversion with minimum` ` ` `// operations.` ` ` `else` `if` `(j == ` `0` `) {` ` ` `dp[i][j] = i;` ` ` `}` ` ` ` ` `else` `{` ` ` `// find the minimum among three` ` ` `// operations below` ` ` ` ` ` ` `dp[i][j] = minm_edits(dp[i - ` `1` `][j - ` `1` `]` ` ` `+ NumOfReplacement(str1.charAt(i - ` `1` `),str2.charAt(j - ` `1` `)), ` `// replace` ` ` `dp[i - ` `1` `][j] + ` `1` `, ` `// delete` ` ` `dp[i][j - ` `1` `] + ` `1` `); ` `// insert` ` ` `}` ` ` `}` ` ` `}` ` ` ` ` `return` `dp[str1.length()][str2.length()];` ` ` `}` ` ` ` ` `// check for distinct characters` ` ` `// in str1 and str2` ` ` ` ` `static` `int` `NumOfReplacement(` `char` `c1, ` `char` `c2)` ` ` `{` ` ` `return` `c1 == c2 ? ` `0` `: ` `1` `;` ` ` `}` ` ` ` ` `// receives the count of different` ` ` `// operations performed and returns the` ` ` `// minimum value among them.` ` ` ` ` `static` `int` `minm_edits(` `int` `... nums)` ` ` `{` ` ` ` ` `return` `Arrays.stream(nums).min().orElse(` ` ` `Integer.MAX_VALUE);` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` ` ` `String s1 = ` `"glomax"` `;` ` ` `String s2 = ` `"folmax"` `;` ` ` ` ` `System.out.println(compute_Levenshtein_distanceDP(s1, s2));` ` ` `}` `}` |

**Output**

3

**Time Complexity: **O(m*n), where m is the length of the first string, and n is the length of the second string.

**Auxiliary Space: **O(m*n), as the matrix used in the above implementation has dimensions **m*n**.

**Applications: **

- Spell Checkers.
- Speech Recognition.
- DNA Analysis.

Attention reader! Don’t stop learning now. Get hold of all the important **Java Foundation** and Collections concepts with the **Fundamentals of Java and Java Collections Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**