1282-group-the-people-given-the-group-size-they-belong-to.rb

# frozen_string_literal: true

# 1282. Group the People Given the Group Size They Belong To
# Medium
# https://leetcode.com/problems/group-the-people-given-the-group-size-they-belong-to

=begin
There are n people that are split into some unknown number of groups. Each person is labeled with a unique ID from 0 to n - 1.
You are given an integer array groupSizes, where groupSizes[i] is the size of the group that person i is in. For example, if groupSizes[1] = 3, then person 1 must be in a group of size 3.
Return a list of groups such that each person i is in a group of size groupSizes[i].
Each person should appear in exactly one group, and every person must be in a group. If there are multiple answers, return any of them. It is guaranteed that there will be at least one valid solution for the given input.

Example 1:
Input: groupSizes = [3,3,3,3,3,1,3]
Output: [[5],[0,1,2],[3,4,6]]
Explanation:
The first group is [5]. The size is 1, and groupSizes[5] = 1.
The second group is [0,1,2]. The size is 3, and groupSizes[0] = groupSizes[1] = groupSizes[2] = 3.
The third group is [3,4,6]. The size is 3, and groupSizes[3] = groupSizes[4] = groupSizes[6] = 3.
Other possible solutions are [[2,1,6],[5],[0,4,3]] and [[5],[0,6,2],[4,3,1]].

Example 2:
Input: groupSizes = [2,1,3,3,3,2]
Output: [[1],[0,5],[2,3,4]]


Constraints:
* groupSizes.length == n
* 1 <= n <= 500
* 1 <= groupSizes[i] <= n
=end

# @param {Integer[]} group_sizes
# @return {Integer[][]}
def group_the_people(group_sizes)
  people = Hash.new() { |h, k| h[k] = Array.new }
  groups = []
  group_sizes.each.with_index { |size, idx| people[size] << idx }
  people.each { |size, ids| groups << ids.pop(size) until ids.empty? }
  groups
end

# **************** #
#       TEST       #
# **************** #

require "test/unit"
class Test_group_the_people < Test::Unit::TestCase
  def test_
    assert_equal [[5], [0, 1, 2], [3, 4, 6]].sort, group_the_people([3, 3, 3, 3, 3, 1, 3]).sort
    assert_equal [[1], [0, 5], [2, 3, 4]].sort, group_the_people([2, 1, 3, 3, 3, 2]).sort
  end
end