Solution Explanation: Minimum Deletion Steps to Make Two Strings Equal

This problem asks for the minimum number of deletions needed to make two strings identical. A dynamic programming approach efficiently solves this.

Approach:

The core idea is to find the longest common subsequence (LCS) of the two input strings. The number of deletions required is simply the sum of the lengths of the two strings minus twice the length of their LCS. This is because the characters in the LCS don't need to be deleted.

Dynamic Programming Table:

We construct a 2D table f where f[i][j] represents the length of the LCS of the first i characters of word1 and the first j characters of word2.

• Base Cases:

  • f[i][0] = 0 for all i (LCS with an empty string is empty)
  • f[0][j] = 0 for all j (LCS with an empty string is empty)

• Recursive Relation:

  • If word1[i-1] == word2[j-1], then f[i][j] = f[i-1][j-1] + 1 (extend the LCS)
  • Otherwise, f[i][j] = max(f[i-1][j], f[i][j-1]) (take the longer LCS from either omitting the last character of word1 or word2)

Code Explanation (Python):

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        f = [[0] * (n + 1) for _ in range(m + 1)]  # Initialize DP table
 
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if word1[i - 1] == word2[j - 1]:
                    f[i][j] = f[i - 1][j - 1] + 1  # Extend LCS
                else:
                    f[i][j] = max(f[i - 1][j], f[i][j - 1])  # Choose max LCS
 
        lcs_length = f[m][n]  # Length of LCS
        return m + n - 2 * lcs_length  # Total deletions

Time and Space Complexity:

• Time Complexity: O(m*n), where 'm' and 'n' are lengths of word1 and word2. This is due to the nested loops filling the DP table.
• Space Complexity: O(m*n), as we use a 2D DP table of size (m+1) x (n+1). This can be optimized to O(min(m,n)) using space optimization techniques (only storing the previous row or column).

The provided solutions in other languages (Java, C++, Go, TypeScript, Rust) follow the same dynamic programming algorithm, differing only in syntax and data structure usage. They all share the same time and space complexity.