Is Partition Array Into K-Distinct Groups easy or hard?

This problem is typically rated Medium because the implementation is simple but requires recognizing the key counting insight. Many candidates initially attempt simulation before realizing the maximum-frequency constraint.

Partition Array Into K-Distinct Groups Python/Java solution

Implement the solution by counting frequencies with a dictionary in Python or a HashMap in Java. Compute the number of groups n/k, track the maximum frequency, and return true if maxFreq <= n/k. The implementation runs in O(n) time.

How to solve Partition Array Into K-Distinct Groups in O(n)?

First compute n/k, which is the number of groups required. Build a frequency map of the array using a hash table. Track the maximum frequency of any element. If maxFreq <= n/k, each occurrence can be placed in a different group, so the partition is valid; otherwise it is impossible.

What is the best approach for Partition Array Into K-Distinct Groups?

The most efficient approach uses frequency counting with a hash map. If the array length is n, you must form n/k groups. Compute the frequency of each value and track the maximum frequency. If the maximum frequency is greater than the number of groups, partitioning is impossible. This runs in O(n) time and O(n) space.

Is Partition Array Into K-Distinct Groups asked at Google/Amazon/Meta?

Problems based on frequency counting, grouping constraints, and hash maps appear frequently in interviews at companies like Amazon, Google, and Meta. Variants that check feasibility of forming groups using frequency limits are common in array and counting interview questions.

What data structure is used in Partition Array Into K-Distinct Groups?

A hash map (or dictionary) is the primary data structure used to store element frequencies. The algorithm relies on counting occurrences and comparing the maximum frequency against the number of required groups.

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

The optimal solution runs in O(n) time because it requires a single pass to build the frequency map and another pass to determine the maximum frequency. Space complexity is O(n) in the worst case when all elements are distinct.

Description Hints 5 Solution Video 15:28

3659. Partition Array Into K-Distinct Groups

Q: How to solve Partition Array Into K-Distinct Groups in O(n)?

First compute n/k, which is the number of groups required. Build a frequency map of the array using a hash table. Track the maximum frequency of any element. If maxFreq <= n/k, each occurrence can be placed in a different group, so the partition is valid; otherwise it is impossible.

Q: What is the best approach for Partition Array Into K-Distinct Groups?

The most efficient approach uses frequency counting with a hash map. If the array length is n, you must form n/k groups. Compute the frequency of each value and track the maximum frequency. If the maximum frequency is greater than the number of groups, partitioning is impossible. This runs in O(n) time and O(n) space.

Q: Is Partition Array Into K-Distinct Groups asked at Google/Amazon/Meta?

Problems based on frequency counting, grouping constraints, and hash maps appear frequently in interviews at companies like Amazon, Google, and Meta. Variants that check feasibility of forming groups using frequency limits are common in array and counting interview questions.

Q: What data structure is used in Partition Array Into K-Distinct Groups?

A hash map (or dictionary) is the primary data structure used to store element frequencies. The algorithm relies on counting occurrences and comparing the maximum frequency against the number of required groups.

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

The optimal solution runs in O(n) time because it requires a single pass to build the frequency map and another pass to determine the maximum frequency. Space complexity is O(n) in the worst case when all elements are distinct.

Medium47.1% acceptance1 company

You are given an integer array nums and an integer k.

Your task is to determine whether it is possible to partition all elements of nums into one or more groups such that:

Each group contains exactly k elements.
All elements in each group are distinct.
Each element in nums must be assigned to exactly one group.

Return true if such a partition is possible, otherwise return false.

Example 1:

Input: nums = [1,2,3,4], k = 2

Output: true

Explanation:

One possible partition is to have 2 groups:

Group 1: [1, 2]
Group 2: [3, 4]

Each group contains k = 2 distinct elements, and all elements are used exactly once.

Example 2:

Input: nums = [3,5,2,2], k = 2

Output: true

Explanation:

One possible partition is to have 2 groups:

Group 1: [2, 3]
Group 2: [2, 5]

Each group contains k = 2 distinct elements, and all elements are used exactly once.

Example 3:

Input: nums = [1,5,2,3], k = 3

Output: false

Explanation:

We cannot form groups of k = 3 distinct elements using all values exactly once.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵
1 <= k <= nums.length

Asked at 1 company

Google

Topics

Hints open one at a time

Hint 1

Think about how many groups of size k you need to form.

Hint 2

Each group must contain exactly k distinct elements.

Hint 3

If any element appears more times than the number of groups groups, it cannot be placed uniquely in each group.

Hint 4

Use a frequency map freq to count the occurrences of each element.

Hint 5

If the total number of elements n is not divisible by k, partitioning is impossible.

15:28

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1class Solution:
2    def partitionArray(self, nums: List[int], k: int) -> bool:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[1,2,3,4]
2

Input

[3,5,2,2]
2

Input

[1,5,2,3]
3

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