# Edit distance 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')


Sol:

Simple DP. Possible to save space.

class Solution {
public:
int minDistance(string word1, string word2) {
int n = word1.size();
int m = word2.size();
if(word1.empty()) return m;
if(word2.empty()) return n;
vector<vector<int>> dp(n+1, vector<int>(m+1,1e9));
for(int i = 0; i < n+1; i++){
for(int j = 0; j < m+1; j++){
if(i == 0) dp[i][j] = j;
else if(j == 0) dp[i][j] = i;
else{
if(word1[i-1] == word2[j-1]){
dp[i][j] = min(dp[i][j], dp[i-1][j-1]);
}else{
dp[i][j] = min(dp[i][j], dp[i-1][j-1]+1);
}
dp[i][j] = min(dp[i][j], dp[i-1][j]+1);
dp[i][j] = min(dp[i][j], dp[i][j-1]+1);
}
}
}
return dp[n][m];
}
};