How do you solve Sort Array by Increasing Frequency on LeetCode?

Sort Array by Increasing Frequency (LeetCode #1636) 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: Counting frequencies is essential for sorting based on occurrences.

What is the brute force solution for Sort Array by Increasing Frequency?

The brute force approach for Sort Array by Increasing Frequency is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we will count the frequency of each number and then sort the array based on these frequencies. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Sort Array by Increasing Frequency?

Sort Array by Increasing Frequency uses the following concepts: Array, Hash Table, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sort Array by Increasing Frequency?

Always clarify the problem requirements and constraints before jumping to coding. Think about edge cases, such as arrays with all unique elements or all identical elements. Practice writing clean and efficient code, and explain your thought process as you code.

What are common mistakes when solving Sort Array by Increasing Frequency?

Forgetting to handle ties in frequency correctly. Not using a stable sorting method, which can affect the order of elements with the same frequency.

#1636

Sort Array by Increasing Frequency

Easy

ArrayHash TableSortingHash MapArray

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 approach uses a frequency map to count occurrences and then sorts the unique numbers based on their frequency and value. This is efficient and leverages sorting effectively.

⚙️

Algorithm

4 steps

1Step 1: Create a frequency map to count occurrences of each number.
2Step 2: Create a list of unique numbers from the original array.
3Step 3: Sort the unique numbers using a custom comparator based on frequency and value.
4Step 4: Build the output array by repeating each number according to its frequency.

solution.py8 lines

1# Full working Python code
2from collections import Counter
3
4def frequencySort(nums):
5    freq = Counter(nums)
6    unique_nums = list(freq.keys())
7    unique_nums.sort(key=lambda x: (freq[x], -x))
8    return [num for num in unique_nums for _ in range(freq[num])]

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step of the unique numbers. The space complexity is O(n) because we store the frequency map and the output array.

1Counting frequencies is essential for sorting based on occurrences.
2Custom sorting logic can handle multiple criteria (frequency and value).

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