Introduction

In this blog post, we will explore how to solve the LeetCode problem "583" 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 steps required to make word1 and word2 the same. In one step, you can delete exactly one character in either string. Example 1: Input: word1 = "sea", word2 = "eat" Output: 2 Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea". Example 2: Input: word1 = "leetcode", word2 = "etco" Output: 4 Constraints: 1 <= word1.length, word2.length <= 500 word1 and word2 consist of only lowercase English letters.

Explanation

Here's a breakdown of the approach, followed by the code:

• Find the Longest Common Subsequence (LCS): The key idea is to find the length of the longest common subsequence present in both strings. This subsequence represents the characters that don't need to be deleted.
• Calculate Deletions: Once we have the LCS length, we can calculate the number of deletions needed in each string. The number of deletions in word1 is len(word1) - LCS_length, and similarly for word2.

• Total Steps: The total number of steps is simply the sum of the deletions needed in both strings.

• Runtime Complexity: O(mn), where 'm' is the length of word1 and 'n' is the length of word2. Storage Complexity: O(mn)

Code

    def minDistance(word1: str, word2: str) -> int:
    """
    Calculates the minimum number of steps (deletions) required to make two words the same.

    Args:
        word1: The first word.
        word2: The second word.

    Returns:
        The minimum number of deletion steps.
    """

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

    # Initialize a DP table to store lengths of LCS
    dp = [[0] * (m + 1) for _ in range(n + 1)]

    # Populate the DP table
    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] + 1
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

    # Length of the Longest Common Subsequence
    lcs_length = dp[n][m]

    # Calculate the number of deletions needed
    deletions_word1 = n - lcs_length
    deletions_word2 = m - lcs_length

    # Return the total number of deletions
    return deletions_word1 + deletions_word2