# Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character
Example

Given word1 = `"mart"` and word2 = `"karma"`, return `3`.

Solution: Dynamic Programming

f[i][j] = MIN(f[i-1][j-1], f[i-1][j]+1, f[i][j-1]+1) // a[i] == b[j]

= MIN(f[i-1][j], f[i][j-1], f[i-1][j-1]) + 1 // a[i] != b[j]

intialize: f[i] = i, f[j] = j

```public int minDistance(String word1, String word2) {
if (word1 == null || word2 == null) {
return -1;
}
//minDistance[i][j] = minimum distance to convert first i chars of word 1
//to first j chars of word 2
int[][] minDistance = new int[word1.length() + 1][word2.length() + 1];
//init the first column -- word2 is empty
for (int i = 0; i < minDistance.length; i++) {
minDistance[i] = i;
}
//init the first row -- word1 is empty
for (int j = 0; j < minDistance.length; j++) {
minDistance[j] = j;
}

for (int i = 1; i <= word1.length(); i++) {
for (int j = 1; j <= word2.length(); j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
minDistance[i][j] = Math.min(Math.min(minDistance[i - 1][j - 1],
minDistance[i - 1][j] + 1), minDistance[i][j - 1] + 1);
} else {
minDistance[i][j] = Math.min(Math.min(minDistance[i - 1][j - 1],
minDistance[i - 1][j]), minDistance[i][j - 1]) + 1;
}
}
}
return minDistance[word1.length()][word2.length()];
}```