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In this post, you will find the solution for the Edit Distance in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.
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Link for the Problem – Edit Distance– LeetCode Problem
Edit Distance– LeetCode Problem
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 <= 500
word1
andword2
consist of lowercase English letters.
Edit Distance– LeetCode Solutions
Edit Distance in C++:
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 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]; } }
Edit Distance 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]