Link: https://leetcode.com/problems/partition-array-into-k-distinct-groups

Problem#

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

Analysis#

The problem revolves around k and whether the elements can be divided evenly into groups. It pays to look for all the failing cases first.

• If the elements cannot be split into groups of exactly k => impossible => early return on that one condition
• If k elements do divide evenly into x groups: because every element within a group has to be unique, this reduces to a counting and distribution problem.
• A hashMap counting occurrences comes to mind immediately — time-complexity: O(n), space-complexity: O(n)

Final solution#

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {boolean}
 */
var partitionArray = function (nums, k) {
    if (nums.length % k !== 0) {
        return false;
    }
    const numOfGroup = nums.length / k;
    const frequency = {};
    for (let item of nums) {
        if (frequency[item]) {
            frequency[item]++;
        }
        else {
            frequency[item] = 1;
        }
        if (frequency[item] > numOfGroup) {
            return false
        }
    }
    return true;
};