How do you solve Partition Array Into K-Distinct Groups on LeetCode?

Partition Array Into K-Distinct Groups (LeetCode #3659) 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: The number of unique elements must be at least k.

What is the time complexity of Partition Array Into K-Distinct Groups?

The optimal solution for Partition Array Into K-Distinct Groups runs in O(n) time and uses O(n) space. Counting frequencies takes linear time, and storing them requires linear space.

What is the brute force solution for Partition Array Into K-Distinct Groups?

The brute force approach for Partition Array Into K-Distinct Groups is Brute Force, with O(n²) time complexity and O(1) space complexity. Try every possible way to group the elements and check if each group meets the requirements. This is 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 Partition Array Into K-Distinct Groups?

Partition Array Into K-Distinct Groups uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Partition Array Into K-Distinct Groups?

Clarify the requirements before coding. Discuss edge cases like empty arrays or all elements being the same. Think about time and space complexity during the solution.

What are common mistakes when solving Partition Array Into K-Distinct Groups?

Assuming duplicates can be used in the same group. Not checking if total elements are divisible by k.

#3659

Partition Array Into K-Distinct Groups

Medium

ArrayHash TableCountingHash 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)

Count the frequency of each element. If the number of unique elements is less than k or if the total number of elements isn't divisible by k, it's impossible to form the groups.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each element in nums.
2Step 2: Check if the number of unique elements is at least k and if the total count of elements is divisible by k.
3Step 3: If both conditions are satisfied, return true; otherwise, return false.

solution.py5 lines

1from collections import Counter
2
3def can_partition(nums, k):
4    freq = Counter(nums)
5    return len(freq) >= k and len(nums) % k == 0

ℹ

Complexity note: Counting frequencies takes linear time, and storing them requires linear space.

1The number of unique elements must be at least k.
2Total elements must be divisible by k.

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