In this post, we are going to solve the Edit Distance Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.
Problem
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
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')
Constraints:
0 <= word1.length, word2.length <= 500word1andword2consist of lowercase English letters.
Now, let’s see the leetcode solution of Edit Distance Leetcode Solution.
Edit Distance Leetcode Solution in Python
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
m = len(word1)
n = len(word2)
# dp[i][j] := min # Of operations to convert word1[0..i) to word2[0..j)
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
dp[i][0] = i
for j in range(1, n + 1):
dp[0][j] = j
for i in range(1, m + 1):
for j in range(1, n + 1):
if word1[i - 1] == word2[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
else:
dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
return dp[m][n]
Edit Distance Leetcode Solution in CPP
class Solution {
public:
int minDistance(string word1, string word2) {
const int m = word1.length();
const int n = word2.length();
// dp[i][j] := min # of operations to convert word1[0..i) to word2[0..j)
vector<vector<int>> dp(m + 1, vector<int>(n + 1));
for (int i = 1; i <= m; ++i)
dp[i][0] = i;
for (int j = 1; j <= n; ++j)
dp[0][j] = j;
for (int i = 1; i <= m; ++i)
for (int j = 1; j <= n; ++j)
if (word1[i - 1] == word2[j - 1])
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
return dp[m][n];
}
};
Edit Distance Leetcode Solution in Java
class Solution {
public int minDistance(String word1, String word2) {
final int m = word1.length();
final int n = word2.length();
// dp[i][j] := min # of operations to convert word1[0..i) to word2[0..j)
int[][] dp = new int[m + 1][n + 1];
for (int i = 1; i <= m; ++i)
dp[i][0] = i;
for (int j = 1; j <= n; ++j)
dp[0][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] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
return dp[m][n];
}
}
Note: This problem Edit Distance is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.
