# Edit Distance in Java

From Wiki:

In computer science, edit distance is a way of quantifying how dissimilar two strings (e.g., words) are to one another by counting the minimum number of operations required to transform one string into the other.

There are three operations permitted on a word: replace, delete, insert. For example, the edit distance between "a" and "b" is 1, the edit distance between "abc" and "def" is 3. This post analyzes how to calculate edit distance by using dynamic programming.

Key Analysis

Let dp[i][j] stands for the edit distance between two strings with length i and j, i.e., word1[0,...,i-1] and word2[0,...,j-1].
There is a relation between dp[i][j] and dp[i-1][j-1]. Let's say we transform from one string to another. The first string has length i and it's last character is "x"; the second string has length j and its last character is "y". The following diagram shows the relation.

1. if x == y, then dp[i][j] == dp[i-1][j-1]
2. if x != y, and we insert y for word1, then dp[i][j] = dp[i][j-1] + 1
3. if x != y, and we delete x for word1, then dp[i][j] = dp[i-1][j] + 1
4. if x != y, and we replace x with y for word1, then dp[i][j] = dp[i-1][j-1] + 1
5. When x!=y, dp[i][j] is the min of the three situations.

Initial condition:
dp[i][0] = i, dp[0][j] = j

Java Code

After the analysis above, the code is just a representation of it.

```public static int minDistance(String word1, String word2) { int len1 = word1.length(); int len2 = word2.length();   // len1+1, len2+1, because finally return dp[len1][len2] int[][] dp = new int[len1 + 1][len2 + 1];   for (int i = 0; i <= len1; i++) { dp[i][0] = i; }   for (int j = 0; j <= len2; j++) { dp[0][j] = j; }   //iterate though, and check last char for (int i = 0; i < len1; i++) { char c1 = word1.charAt(i); for (int j = 0; j < len2; j++) { char c2 = word2.charAt(j);   //if last two chars equal if (c1 == c2) { //update dp value for +1 length dp[i + 1][j + 1] = dp[i][j]; } else { int replace = dp[i][j] + 1; int insert = dp[i][j + 1] + 1; int delete = dp[i + 1][j] + 1;   int min = replace > insert ? insert : replace; min = delete > min ? min : delete; dp[i + 1][j + 1] = min; } } }   return dp[len1][len2]; }```
Category >> Algorithms
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• PrasAnth Mamakar Komaragiri

Hey can you please tell me how to print the matrix at every step in this example

• RoganDawes

True enough. Good point. Simply update the single row in place as you work through it.

• 1 row and 1 value are enough ðŸ™‚

• Bhudev

Yes it should, works either way

• Internet Hero

int replace = dp[i][j] + 1;
int insert = dp[i][j + 1] + 1;
int delete = dp[i + 1][j] + 1;

Shouldn’t this be reverse, ie

insert = dp[i+1][j]+1
delete = dp[i][j+1] + 1

• Vimukthi Weerasiri

for (int i = 1; i < dp.length; i++) {
for (int j = 1; j < dp[0].length; j++) {
dp[i][j] = word2.charAt(i – 1) == word1.charAt(j – 1) ? dp[i – 1][j – 1] : Math.min(dp[i – 1][j – 1], Math.min(dp[i – 1][j], dp[i][j – 1])) + 1;
}
}

A simpler loop which gives the solution

• AThomasTran

Appreciated! This example was *exactly* what I was looking for to solidify my understanding of Min Edit Distance DP problem

• RoganDawes

Note that there is no need to keep the full 2 dimensional array of results. This obviously grows with the product of the length of the two strings. All you really need to keep is 2 rows, and you simply switch them as you move from row to row.

Unfortunately, that doesn’t change the number of operations required to calculate the edit distance, which remains O(m*n), where m and n are the lengths of the strings.