LeetCode 583: Delete Operation for Two Strings Solution in Python – A Step-by-Step Guide

What Is LeetCode 583: Delete Operation for Two Strings?

In LeetCode 583: Delete Operation for Two Strings, you’re given two strings word1 and word2, and your task is to return the minimum number of characters that must be deleted from both strings to make them identical. Each deletion removes one character from either string, and the goal is to find the fewest total deletions. For example, with word1 = "sea" and word2 = "eat", the minimum deletions is 2 (delete 's' from "sea" and 't' from "eat" to get "ea"). This problem builds on LeetCode 1143: Longest Common Subsequence for sequence matching but focuses on minimizing deletions to achieve equality.

Problem Statement

• Input: word1 (str)—first string; word2 (str)—second string.
• Output: Integer—minimum number of deletions to make strings identical.
• Rules: Delete one character per operation from either string; minimize total deletions.

Constraints

• 1 <= word1.length, word2.length <= 10⁵
• word1 and word2 consist of lowercase English letters.

Examples

• Input: word1 = "sea", word2 = "eat"
• Output: 2
• Delete 's' from "sea" (→ "ea"), 't' from "eat" (→ "ea"), total = 2.
• Input: word1 = "leetcode", word2 = "etco"
• Output: 4
• Delete 'l', 'e', 'd', 'e' from "leetcode" (→ "etco"), total = 4.
• Input: word1 = "a", word2 = "b"
• Output: 2
• Delete 'a' and 'b', total = 2.

Understanding the Problem: Minimizing Deletions

To solve LeetCode 583: Delete Operation for Two Strings in Python, we need to find the minimum number of character deletions from word1 and word2 to make them identical, handling strings up to 10⁵ characters efficiently. A brute-force approach trying all deletion combinations (O(2^n)) is impractical. Instead, a key insight—deleting all characters not in the longest common subsequence (LCS)—leads to an efficient dynamic programming solution in O(m * n) time, where m and n are the string lengths. We’ll explore:

• Best Solution (Dynamic Programming with LCS): O(m n) time, O(m n) space—fast and optimal.
• Alternative Solution (Brute-Force Recursive Deletion): O(2^(m+n)) time, O(m + n) space—thorough but slow.

Let’s dive into the DP solution with a friendly breakdown!

Best Solution: Dynamic Programming with Longest Common Subsequence

Why DP Wins

The dynamic programming with LCS solution is the best for LeetCode 583 because it computes the minimum deletions in O(m * n) time and O(m * n) space by finding the longest common subsequence between word1 and word2, then calculating deletions as the total length minus twice the LCS length, efficiently avoiding exhaustive deletion checks. It’s like finding the common ground between two strings and trimming the excess—all in a structured, optimized way!

How It Works

Think of this as a string-alignment strategist:

• Step 1: Understand Deletion Goal:
• Min deletions = len(word1) + len(word2) - 2 * LCS length.
• Step 2: Build DP Table:
• dp[i][j]: LCS length of word1[:i] and word2[:j].
• Step 3: Fill Table:
• If characters match: dp[i][j] = dp[i-1][j-1] + 1.
• If not: dp[i][j] = max(dp[i-1][j], dp[i][j-1]).
• Step 4: Compute Deletions:
• Use final LCS length to calculate deletions.
• Step 5: Return Result:
• Minimum number of deletions.
• Why It Works:
• LCS identifies characters to keep; rest are deleted.
• DP avoids recomputing subproblems.

It’s like a deletion-minimizing maestro!

Step-by-Step Example

Example: word1 = "sea", word2 = "eat"

• Step 1: Lengths: m = 3, n = 3, goal = 3 + 3 - 2 * LCS.
• Step 2: DP table (4x4, including empty prefixes):
• Initialize: dp[0][j] = 0, dp[i][0] = 0.
``` "" e a t "" 0 0 0 0 s 0 0 0 0 e 0 1 1 1 a 0 1 2 2 ```
• Fill:
• (1,1): 's' ≠ 'e', max(0, 0) = 0.
• (2,1): 'e' = 'e', dp[1][0] + 1 = 1.
• (3,2): 'a' = 'a', dp[2][1] + 1 = 2.
• Final: dp[3][3] = 2 (LCS = "ea").
• Step 3: Deletions = 3 + 3 - 2 * 2 = 2.
• Result: 2.

Example: word1 = "ab", word2 = "cd"

• Step 2: DP table:
``` "" c d "" 0 0 0 a 0 0 0 b 0 0 0 ```
• Final: dp[2][2] = 0 (no common subsequence).
• Step 3: Deletions = 2 + 2 - 2 * 0 = 4.
• Result: 4.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)

        # Step 1: Initialize DP table
        dp = [[0] * (n + 1) for _ in range(m + 1)]

        # Step 2: Fill DP table for LCS
        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] + 1
                else:
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

        # Step 3: Compute minimum deletions
        lcs_length = dp[m][n]
        return m + n - 2 * lcs_length

• Line 4: Get lengths of both strings.
• Line 7: Create (m+1) x (n+1) DP table with zeros.
• Lines 10-14: Fill table:
• Match: Add 1 to diagonal (previous LCS + 1).
• No match: Take max of left or up (longer LCS excluding current char).
• Lines 17-18: Compute deletions using LCS length.
• Time Complexity: O(m * n)—fill DP table.
• Space Complexity: O(m * n)—DP table storage.

It’s like a string-trimming genius!

Alternative Solution: Brute-Force Recursive Deletion

Why an Alternative Approach?

The brute-force recursive deletion approach tries all possible deletion combinations from both strings to find the minimum needed to make them equal, running in O(2^(m+n)) time and O(m + n) space due to recursion depth. It’s thorough but impractical for large strings, making it a good baseline for small inputs or understanding.

How It Works

Picture this as a deletion-exploring seeker:

• Step 1: Define recursive function with indices.
• Step 2: Base case: If strings equal, return 0.
• Step 3: Recursive case:
• If chars match, move both indices.
• Else, try deleting from either string, take min.
• Step 4: Return minimum deletions.

It’s like a deletion-path adventurer!

Step-by-Step Example

Example: word1 = "ab", word2 = "a"

• Step 1: Call delete(0, 0):
• 'a' = 'a', delete(1, 1):
• "b" vs "", try:
• Delete 'b': delete(2, 1) → "" = "", 0 + 1 = 1.
• Step 2: Min deletions = 1.
• Result: 1.

Code for Brute-Force Approach

Copy codeclass Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        def delete(i: int, j: int) -> int:
            # Base case: one string exhausted
            if i == len(word1):
                return len(word2) - j
            if j == len(word2):
                return len(word1) - i

            # Recursive case
            if word1[i] == word2[j]:
                return delete(i + 1, j + 1)
            return 1 + min(delete(i + 1, j), delete(i, j + 1))

        return delete(0, 0)

• Lines 4-8: Base: Return remaining length if one string ends.
• Lines 11-13: Recurse:
• Match: Move both pointers.
• No match: Delete from either, add 1, take min.
• Line 15: Start from (0, 0).
• Time Complexity: O(2^(m+n))—exponential combinations.
• Space Complexity: O(m + n)—recursion stack.

It’s a brute-force deletion tester!

Comparing the Two Solutions

• DP (Best):
• Pros: O(m n), O(m n), fast.
• Cons: Space for DP table.
• Brute-Force (Alternative):
• Pros: O(2^(m+n)), O(m + n), thorough.
• Cons: Impractical for large strings.

DP wins for efficiency!

Additional Examples and Edge Cases

• "a", "a": 0.
• "abc", "def": 6.
• Empty strings: 0.

DP handles them all!

Complexity Recap

• DP: Time O(m n), Space O(m n).
• Brute-Force: Time O(2^(m+n)), Space O(m + n).

DP’s the speed champ!

Key Takeaways

• DP: LCS magic—learn at Python Basics!
• Brute-Force: Deletion roam.
• Strings: Matching is fun.
• Python: DP or recurse, your pick!

Final Thoughts: Trim Those Strings!

LeetCode 583: Delete Operation for Two Strings in Python is a clever DP challenge. Dynamic programming with LCS is your fast track, while brute-force offers a thorough dive. Want more? Try LeetCode 1143: Longest Common Subsequence or LeetCode 72: Edit Distance. Ready to delete? Head to Solve LeetCode 583 on LeetCode and minimize those deletions today!