Introduction

In this blog post, we will explore how to solve the LeetCode problem "72" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the breakdown of the optimal approach, complexity analysis, and the Python code for solving the Edit Distance problem:

• Dynamic Programming: Use dynamic programming to build a table where dp[i][j] represents the minimum edit distance between the first i characters of word1 and the first j characters of word2.
• Base Cases and Recurrence Relation: Initialize the first row and column of the DP table to represent the edit distance when one of the strings is empty. The core recurrence relation considers insertion, deletion, and replacement operations to determine the minimum cost.

• Optimal Substructure: The solution leverages the optimal substructure property, where the optimal solution for larger subproblems can be constructed from the optimal solutions of smaller, overlapping subproblems.

• Time Complexity: O(mn), where m and n are the lengths of word1 and word2 respectively. Space Complexity: O(mn)

Code

    def minDistance(word1: str, word2: str) -> int:
    """
    Calculates the minimum edit distance between two words using dynamic programming.

    Args:
        word1: The first word (string).
        word2: The second word (string).

    Returns:
        The minimum number of operations required to convert word1 to word2.
    """

    n = len(word1)
    m = len(word2)

    # Initialize a DP table with dimensions (n+1) x (m+1)
    dp = [[0] * (m + 1) for _ in range(n + 1)]

    # Base cases: If word1 is empty, the edit distance is the length of word2 (insertions)
    for j in range(m + 1):
        dp[0][j] = j

    # Base cases: If word2 is empty, the edit distance is the length of word1 (deletions)
    for i in range(n + 1):
        dp[i][0] = i

    # Fill the DP table using the recurrence relation
    for i in range(1, n + 1):
        for j in range(1, m + 1):
            if word1[i - 1] == word2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]  # No operation needed
            else:
                dp[i][j] = 1 + min(
                    dp[i - 1][j],  # Deletion
                    dp[i][j - 1],  # Insertion
                    dp[i - 1][j - 1],  # Replacement
                )

    # The minimum edit distance is at dp[n][m]
    return dp[n][m]