How do you solve Sort Characters By Frequency on LeetCode?

Sort Characters By Frequency (LeetCode #451) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using a frequency map is essential for counting occurrences efficiently.

What is the brute force solution for Sort Characters By Frequency?

The brute force approach for Sort Characters By Frequency is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves counting the frequency of each character and then sorting the characters based on their frequencies. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Sort Characters By Frequency?

Explain your thought process clearly while coding. Consider edge cases like empty strings or strings with all unique characters. Practice writing clean and efficient code.

What are common mistakes when solving Sort Characters By Frequency?

Not considering character case sensitivity. Forgetting to handle ties in frequency correctly.

#451

Sort Characters By Frequency

Medium

Hash TableStringSortingHeap (Priority Queue)Bucket SortCountingHash MapHeap

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution uses a frequency map and a max heap (priority queue) to efficiently sort characters by their frequency. This approach is faster and scales better with larger inputs.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each character using a HashMap.
2Step 2: Use a max heap to store characters based on their frequencies.
3Step 3: Construct the result string by repeatedly extracting characters from the heap.

solution.py13 lines

1# Full working Python code
2import heapq
3from collections import Counter
4
5def frequencySort(s):
6    freq = Counter(s)
7    max_heap = [(-count, char) for char, count in freq.items()]
8    heapq.heapify(max_heap)
9    result = []
10    while max_heap:
11        count, char = heapq.heappop(max_heap)
12        result.append(char * -count)
13    return ''.join(result)

ℹ

Complexity note: The time complexity is O(n log n) due to the heap operations for sorting the characters. The space complexity is O(n) because we store the frequency of each character.

1Using a frequency map is essential for counting occurrences efficiently.
2Max heaps are useful for sorting elements based on custom criteria.

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