How do you solve Rearrange Characters to Make Target String on LeetCode?

Rearrange Characters to Make Target String (LeetCode #2287) 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: Counting character frequencies is crucial for solving this problem efficiently.

What is the brute force solution for Rearrange Characters to Make Target String?

The brute force approach for Rearrange Characters to Make Target String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of characters from string `s` to see how many times we can form the `target` string. This is simple but inefficient, as it can take a long time for larger strings. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Rearrange Characters to Make Target String?

Rearrange Characters to Make Target String uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Rearrange Characters to Make Target String?

Always clarify if characters can be reused or not before starting your solution. Discuss your thought process and approach before jumping into coding. Consider edge cases, such as when `s` has fewer characters than `target`.

What are common mistakes when solving Rearrange Characters to Make Target String?

Not considering the frequency of each character in both strings. Assuming characters can be reused when they cannot.

#2287

Rearrange Characters to Make Target String

Easy

Hash TableStringCountingHash MapArray

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 counts the frequency of each character in both `s` and `target`, then calculates how many times we can form `target` based on the available characters in `s`. This is efficient and avoids unnecessary combinations.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each character in `s` and `target`.
2Step 2: For each character in `target`, calculate how many times we can use the characters from `s` to form `target`.
3Step 3: The result is the minimum number of times we can form `target` based on the limiting character.

solution.py9 lines

1from collections import Counter
2
3def maxCopiesOptimal(s, target):
4    s_count = Counter(s)
5    target_count = Counter(target)
6    copies = float('inf')
7    for char in target_count:
8        copies = min(copies, s_count[char] // target_count[char])
9    return copies

ℹ

Complexity note: The time complexity is O(n + m) because we traverse both strings once to count characters, where n is the length of `s` and m is the length of `target`. The space complexity is O(n + m) due to the storage of character counts.

1Counting character frequencies is crucial for solving this problem efficiently.
2The limiting factor for forming the target string is the character that appears the least number of times in `s` relative to its requirement in `target`.

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