How do you solve Custom Sort String on LeetCode?

Custom Sort String (LeetCode #791) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding character frequency is crucial for optimal solutions.

What is the brute force solution for Custom Sort String?

The brute force approach for Custom Sort String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible permutations of the string `s` and checking which ones match the order defined in `order`. This is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Custom Sort String?

Custom Sort String uses the following concepts: Hash Table, String, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Custom Sort String?

Explain your thought process clearly when discussing your approach. Be ready to discuss time and space complexity. Practice writing clean and efficient code.

What are common mistakes when solving Custom Sort String?

Overlooking characters in `s` that are not in `order`. Not considering the order of characters in the final output.

#791

Custom Sort String

Medium

Hash TableStringSortingHash 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 a frequency count of characters in `s` and constructs the result string based on the order defined in `order`. This is efficient because it processes each character only a couple of times.

⚙️

Algorithm

4 steps

1Step 1: Count the frequency of each character in `s` using a HashMap.
2Step 2: Initialize an empty result string.
3Step 3: Append characters from `order` to the result string based on their frequency in `s`.
4Step 4: Append any remaining characters from `s` that are not in `order` to the result string.

solution.py12 lines

1from collections import Counter
2
3def customSortString(order, s):
4    count = Counter(s)
5    result = []
6    for char in order:
7        if char in count:
8            result.append(char * count[char])
9            del count[char]
10    for char in count:
11        result.append(char * count[char])
12    return ''.join(result)

ℹ

Complexity note: The time complexity is O(n) because we traverse the string `s` to count characters and then again to build the result string. The space complexity is O(n) due to the storage of character frequencies.

1Understanding character frequency is crucial for optimal solutions.
2Using a HashMap allows for quick lookups and efficient counting.

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