# Question

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

1. Insert a character
2. Delete a character
3. Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')


Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')


# Solution

This is a classic problem of Dynamic Programming. We define the state dp[i][j] to be the minimum number of operations to convert word1[0..i - 1] to word2[0..j - 1]. The state equations have two cases: the boundary case and the general case. Note that in the above notations, both i and j take values starting from 1 as we reserve 0 for empty strings.

For the boundary case, that is, to convert a string to an empty string, it is easy to see that the minimum number of operations to convert word1[0..i - 1] to "" requires at least i operations (deletions). In fact, the boundary case is simply:

1. dp[i] = i;
2. dp[j] = j.

Now let’s move on to the general case, that is, convert a non-empty word1[0..i - 1] to another non-empty word2[0..j - 1]. Well, let’s try to break this problem down into smaller problems (sub-problems). Suppose we have already known how to convert word1[0..i - 2] to word2[0..j - 2], which is dp[i - 1][j - 1]. Now let’s consider word[i - 1] and word2[j - 1]. If they are equal, then no more operation is needed and dp[i][j] = dp[i - 1][j - 1]. Well, what if they are not equal?

If they are not equal, we need to consider three cases:

1. Replace word1[i - 1] by word2[j - 1] (dp[i][j] = dp[i - 1][j - 1] + 1 (for replacement));
2. Delete word1[i - 1] and word1[0..i - 2] = word2[0..j - 1] (dp[i][j] = dp[i - 1][j] + 1 (for deletion));
3. Insert word2[j - 1] to word1[0..i - 1] and word1[0..i - 1] + word2[j - 1] = word2[0..j - 1] (dp[i][j] = dp[i][j - 1] + 1 (for insertion)).

Make sure you understand the subtle differences between the equations for deletion and insertion. For deletion, we are actually converting word1[0..i - 2] to word2[0..j - 1], which costs dp[i - 1][j], and then deleting the word1[i - 1], which costs 1. The case is similar for insertion.

Putting these together, we now have:

1. dp[i] = i;
2. dp[j] = j;
3. dp[i][j] = dp[i - 1][j - 1], if word1[i - 1] = word2[j - 1];
4. dp[i][j] = min(dp[i - 1][j - 1] + 1, dp[i - 1][j] + 1, dp[i][j - 1] + 1), otherwise.
class Solution {
public int minDistance(String word1, String word2) {
int m = word1.length(), n = word2.length();

int[][] dp = new int[m + 1][n + 1];

for(int i = 0; i <= m; i++) {
dp[i] = i;
}
for(int j = 0; j <= n; j++) {
dp[j] = j;
}

for(int i = 1; i <= m; i++) {
for(int j = 1; j <= n; j++) {
if(word1.charAt(i - 1) == word2.charAt(j - 1))
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = min(dp[i - 1][j - 1] + 1, dp[i - 1][j] + 1, dp[i][j - 1] + 1);
}
}
return dp[m][n];
}

private int min(int a, int b, int c) {
if(a <= b && a <= c) return a;
else if (b <= a && b <= c) return b;
else return c;
}
}


Well, you may have noticed that each time when we update dp[i][j], we only need dp[i - 1][j - 1], dp[i][j - 1], dp[i - 1][j]. In fact, we need not maintain the full m*n matrix. Instead, maintaining one column is enough. The code can be optimized to $$O(m)$$ or $$O(n)$$ space, depending on whether you maintain a row or a column of the original matrix.

class Solution {
public int minDistance(String word1, String word2) {
int m = word1.length(), n = word2.length();

int[] dp = new int[n + 1];

for(int j = 0; j <= n; j++) {
dp[j] = j;
}

for(int i = 1; i <= m; i++) {
int[] newDp = new int[n + 1];
newDp = i;
for(int j = 1; j <= n; j++) {
if(word1.charAt(i - 1) == word2.charAt(j - 1))
newDp[j] = dp[j - 1];
else
newDp[j] = min(dp[j - 1] + 1, dp[j] + 1, newDp[j - 1] + 1);
}
dp = newDp;
}
return dp[n];
}

private int min(int a, int b, int c) {
if(a <= b && a <= c) return a;
else if (b <= a && b <= c) return b;
else return c;
}
}


The time completely is $$O(mn)$$.