What is the brute force solution for Minimum Number of Steps to Make Two Strings Anagram II?

The brute force approach for Minimum Number of Steps to Make Two Strings Anagram II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves comparing each character in both strings and counting how many characters need to be added to make them anagrams. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Steps to Make Two Strings Anagram II?

Minimum Number of Steps to Make Two Strings Anagram II uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Number of Steps to Make Two Strings Anagram II?

Always think about how to count occurrences efficiently. Discuss your thought process clearly, especially when it comes to edge cases. Practice similar problems to recognize patterns in string manipulation.

What are common mistakes when solving Minimum Number of Steps to Make Two Strings Anagram II?

Not considering all unique characters from both strings. Forgetting to account for characters that are only in one string.

#2186

Minimum Number of Steps to Make Two Strings Anagram II

Medium

Hash TableStringCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses character frequency counts to determine how many characters need to be added to each string. This approach is efficient because it only requires a single pass through each string to count characters, followed by a comparison of counts.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each character in both strings using a HashMap.
2Step 2: For each unique character in both strings, calculate the absolute difference in counts.
3Step 3: Sum these differences to get the total number of steps required.

solution.py13 lines

1# Full working Python code
2from collections import Counter
3
4def min_steps(s, t):
5    count_s = Counter(s)
6    count_t = Counter(t)
7    steps = 0
8    for char in set(count_s.keys()).union(set(count_t.keys())):
9        steps += abs(count_s[char] - count_t[char])
10    return steps
11
12# Example usage
13print(min_steps('leet tco de', 'co a t s'))  # Output: 7

ℹ

Complexity note: This complexity is linear because we only traverse each string once to count characters, and then we perform a constant-time operation for each unique character.

1Character frequency is key to determining how many characters need to be added.
2Using a HashMap allows efficient counting and comparison of character frequencies.

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