Edit Distance LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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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 all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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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About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either EasyMedium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

  • Mathematics/Basic Logical Based Questions
  • Arrays
  • Strings
  • Hash Table
  • Dynamic Programming
  • Stack & Queue
  • Trees & Graphs
  • Greedy Algorithms
  • Breadth-First Search
  • Depth-First Search
  • Sorting & Searching
  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the ProblemEdit 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 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]

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