Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.
You have the following three operations permitted on a word:
- Insert a character
- Delete a character
- Replace a character
Test Cases
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
class Solution {
public:
int minDistance(string word1, string word2) {
int n1 = word1.length(), n2 = word2.length();
vector<vector<int>> dp(n1+1, vector<int>(n2+1, 0));
for (int i = 1; i <= n1; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= n2; j++) {
dp[0][j] = j;
}
for(int i=1; i<=n1; i++) {
for(int j=1; j<=n2; j++) {
if (word1[i-1] != word2[j-1]) {
dp[i][j] = 1 + min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1]));
} else {
dp[i][j] = dp[i-1][j-1];
}
}
}
return dp[n1][n2];
}
};class Solution {
public int minDistance(String word1, String word2) {
int n1 = word1.length(), n2 = word2.length();
int[][] dp = new int[n1+1][n2+1];
for (int i = 1; i <= n1; i++) {
dp[i][0] = i;
}
for (int j = 1; j <= n2; j++) {
dp[0][j] = j;
}
for(int i=1; i<=n1; i++) {
for(int j=1; j<=n2; j++) {
if (word1.charAt(i-1) != word2.charAt(j-1)) {
dp[i][j] = 1 + Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1]));
} else {
dp[i][j] = dp[i-1][j-1];
}
}
}
return dp[n1][n2];
}
}