How do you solve Minimum Steps to Convert String with Operations on LeetCode?

Minimum Steps to Convert String with Operations (LeetCode #3579) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n²) space complexity. Key insight: Transformations can be optimized by considering both original and reversed substrings.

What is the time complexity of Minimum Steps to Convert String with Operations?

The optimal solution for Minimum Steps to Convert String with Operations runs in O(n²) time and uses O(n²) space. The DP approach requires a 2D array to store results for all substring pairs, leading to quadratic space and time complexity.

What is the brute force solution for Minimum Steps to Convert String with Operations?

The brute force approach for Minimum Steps to Convert String with Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach tries every possible way to split the string and applies all operations to transform word1 into word2. It’s simple but inefficient due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Minimum Steps to Convert String with Operations?

Minimum Steps to Convert String with Operations uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, String Manipulation.

What are the interview tips for Minimum Steps to Convert String with Operations?

Clarify the problem constraints and examples. Discuss your thought process before coding. Test edge cases during the interview.

What are common mistakes when solving Minimum Steps to Convert String with Operations?

Not considering all operations for each substring. Ignoring the effect of reversing substrings.

#3579

Minimum Steps to Convert String with Operations

Hard

StringDynamic ProgrammingDynamic ProgrammingString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n²)Space O(n²)

This approach uses dynamic programming to efficiently calculate the minimum operations needed by considering both the original and reversed substrings, thus reducing redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i][j] represents the minimum operations to convert word1[0:i] to word2[0:j].
2Step 2: Fill the DP array by considering replacing, swapping, and reversing substrings.
3Step 3: Return dp[n][n] where n is the length of the strings.

solution.py11 lines

1def min_steps(word1, word2):
2    n = len(word1)
3    dp = [[0] * (n + 1) for _ in range(n + 1)]
4    for i in range(1, n + 1):
5        dp[i][0] = i
6    for j in range(1, n + 1):
7        dp[0][j] = j
8    for i in range(1, n + 1):
9        for j in range(1, n + 1):
10            dp[i][j] = dp[i-1][j-1] + (word1[i-1] != word2[j-1])
11    return dp[n][n]

ℹ

Complexity note: The DP approach requires a 2D array to store results for all substring pairs, leading to quadratic space and time complexity.

1Transformations can be optimized by considering both original and reversed substrings.
2Dynamic programming helps avoid redundant calculations.

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