How do you solve Delete Operation for Two Strings on LeetCode?

Delete Operation for Two Strings (LeetCode #583) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * m) space complexity. Key insight: Understanding the concept of the longest common subsequence (LCS) is crucial for solving this problem efficiently.

What is the brute force solution for Delete Operation for Two Strings?

The brute force approach for Delete Operation for Two Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible ways to delete characters from both strings until they match. This method is straightforward but inefficient, as it explores all combinations of deletions. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Delete Operation for Two Strings?

Delete Operation for Two Strings uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Longest Common Subsequence.

What are the interview tips for Delete Operation for Two Strings?

Always clarify the problem requirements and constraints before diving into coding. Explain your thought process and approach clearly to the interviewer as you code. Practice dynamic programming problems to become comfortable with the concepts and patterns.

What are common mistakes when solving Delete Operation for Two Strings?

Failing to correctly initialize the DP table, leading to incorrect results. Not considering edge cases, such as when one string is empty.

#583

Delete Operation for Two Strings

Medium

StringDynamic ProgrammingDynamic ProgrammingLongest Common Subsequence

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n * m)

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Intuition

Time O(n * m)Space O(n * m)

The optimal approach uses dynamic programming to find the longest common subsequence (LCS) of the two strings. The minimum number of deletions required is the sum of the lengths of both strings minus twice the length of the LCS.

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Algorithm

3 steps

1Step 1: Create a 2D array dp where dp[i][j] represents the length of the LCS of the first i characters of word1 and the first j characters of word2.
2Step 2: Fill the dp array using the following rules: - If characters match (word1[i-1] == word2[j-1]), then dp[i][j] = dp[i-1][j-1] + 1. - If they do not match, then dp[i][j] = max(dp[i-1][j], dp[i][j-1]).
3Step 3: The result is calculated as (len(word1) + len(word2) - 2 * dp[len(word1)][len(word2)]).

solution.py15 lines

1# Full working Python code
2
3def min_steps_optimal(word1, word2):
4    m, n = len(word1), len(word2)
5    dp = [[0] * (n + 1) for _ in range(m + 1)]
6
7    for i in range(1, m + 1):
8        for j in range(1, n + 1):
9            if word1[i - 1] == word2[j - 1]:
10                dp[i][j] = dp[i - 1][j - 1] + 1
11            else:
12                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
13
14    lcs_length = dp[m][n]
15    return (m + n - 2 * lcs_length)

ℹ

Complexity note: The time complexity is O(n * m) because we fill a 2D array where n and m are the lengths of the two strings. The space complexity is also O(n * m) due to the storage of the dp array.

1Understanding the concept of the longest common subsequence (LCS) is crucial for solving this problem efficiently.
2Dynamic programming is often used for problems involving sequences and subsequences, as it allows us to build solutions incrementally.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.