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Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

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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:3Explanation:horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e')

**Example 2:**

Input:word1 = "intention", word2 = "execution"Output:5Explanation: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`

and`word2`

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]