LeetCode #3659 — MEDIUM

Partition Array Into K-Distinct Groups

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[1,2,3,4]
2

Example 2

[3,5,2,2]
2

Example 3

[1,5,2,3]
3

Step 02

Core Insight

What unlocks the optimal approach

Think about how many groups of size <code>k</code> you need to form.
Each group must contain exactly <code>k</code> distinct elements.
If any element appears more times than the number of groups <code>groups</code>, it cannot be placed uniquely in each group.
Use a frequency map <code>freq</code> to count the occurrences of each element.
If the total number of elements <code>n</code> is not divisible by <code>k</code>, partitioning is impossible.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3659: Partition Array Into K-Distinct Groups
class Solution {
    public boolean partitionArray(int[] nums, int k) {
        int n = nums.length;
        if (n % k != 0) {
            return false;
        }
        int m = n / k;
        int mx = Arrays.stream(nums).max().getAsInt();
        int[] cnt = new int[mx + 1];
        for (int x : nums) {
            if (++cnt[x] > m) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #3659: Partition Array Into K-Distinct Groups
func partitionArray(nums []int, k int) bool {
	n := len(nums)
	if n%k != 0 {
		return false
	}
	m := n / k
	mx := slices.Max(nums)
	cnt := make([]int, mx+1)
	for _, x := range nums {
		if cnt[x]++; cnt[x] > m {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #3659: Partition Array Into K-Distinct Groups
class Solution:
    def partitionArray(self, nums: List[int], k: int) -> bool:
        m, mod = divmod(len(nums), k)
        if mod:
            return False
        return max(Counter(nums).values()) <= m

// Accepted solution for LeetCode #3659: Partition Array Into K-Distinct Groups
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3659: Partition Array Into K-Distinct Groups
// class Solution {
//     public boolean partitionArray(int[] nums, int k) {
//         int n = nums.length;
//         if (n % k != 0) {
//             return false;
//         }
//         int m = n / k;
//         int mx = Arrays.stream(nums).max().getAsInt();
//         int[] cnt = new int[mx + 1];
//         for (int x : nums) {
//             if (++cnt[x] > m) {
//                 return false;
//             }
//         }
//         return true;
//     }
// }

// Accepted solution for LeetCode #3659: Partition Array Into K-Distinct Groups
function partitionArray(nums: number[], k: number): boolean {
    const n = nums.length;
    if (n % k) {
        return false;
    }
    const m = n / k;
    const mx = Math.max(...nums);
    const cnt: number[] = Array(mx + 1).fill(0);
    for (const x of nums) {
        if (++cnt[x] > m) {
            return false;
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.