What is the time complexity of Minimum Number of Steps to Make Two Strings Anagram?

The optimal solution for Minimum Number of Steps to Make Two Strings Anagram runs in O(n) time and uses O(1) space. The time complexity is linear because we only traverse each string once to count characters, making it efficient for large inputs.

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

The brute force approach for Minimum Number of Steps to Make Two Strings Anagram is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible character replacement in string t to see if we can make it an anagram of string s. This is simple but inefficient as it may require many iterations. 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?

Minimum Number of Steps to Make Two Strings Anagram 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?

Always clarify the problem constraints and requirements before diving into coding. Think about edge cases, such as strings that are already anagrams. Discuss your thought process with the interviewer as you work through the problem.

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

Not accounting for characters that are present in one string but not the other. Failing to initialize frequency arrays correctly.

#1347

Minimum Number of Steps to Make Two Strings Anagram

Medium

Hash TableStringCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages frequency counting to determine how many characters need to be replaced in t to match the character counts in s. This is efficient and avoids unnecessary iterations.

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Algorithm

4 steps

1Step 1: Create two frequency arrays for characters in s and t.
2Step 2: Count the frequency of each character in both strings.
3Step 3: For each character, calculate the difference in frequency between s and t. If t has fewer of a character than s, add that difference to the total steps needed.
4Step 4: Return the total number of steps.

solution.py12 lines

1def min_steps(s, t):
2    count_s = [0] * 26
3    count_t = [0] * 26
4    for char in s:
5        count_s[ord(char) - ord('a')] += 1
6    for char in t:
7        count_t[ord(char) - ord('a')] += 1
8    steps = 0
9    for i in range(26):
10        if count_t[i] < count_s[i]:
11            steps += count_s[i] - count_t[i]
12    return steps

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Complexity note: The time complexity is linear because we only traverse each string once to count characters, making it efficient for large inputs.

1Character frequency is key to determining the number of replacements needed.
2Understanding how to compare character counts efficiently can save time.

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