1<<-DOC
 2Find the kth largest element in an unsorted array. This will be the kth largest element in sorted order, not the kth distinct element.
 3
 4Example
 5
 6For nums = [7, 6, 5, 4, 3, 2, 1] and k = 2, the output should be
 7kthLargestElement(nums, k) = 6;
 8For nums = [99, 99] and k = 1, the output should be
 9kthLargestElement(nums, k) = 99.
10Input/Output
11
12[time limit] 4000ms (rb)
13[input] array.integer nums
14
15Guaranteed constraints:
161 ≤ nums.length ≤ 105,
17-105 ≤ nums[i] ≤ 105.
18
19[input] integer k
20
21Guaranteed constraints:
221 ≤ k ≤ nums.length.
23
24[output] integer
25DOC
26
27describe "#kth_largest_element" do
28  def kth_largest_element(numbers, k)
29    numbers.sort[-k]
30  end
31
32  #def kth_largest_element(numbers, k)
33    #items = Array.new(105)
34    #numbers.each do |n|
35      #items[n] = n
36    #end
37    #items.compact[-k]
38  #end
39
40  def kth_largest_element(numbers, k)
41    return numbers[0] if numbers.size == 1
42
43    numbers.shuffle!
44    partition = numbers[0]
45    upper = numbers[1..-1].find_all { |x| x >= partition }
46
47    if upper.size >= k
48      kth_largest_element(upper, k)
49    else
50      lower = numbers[1..-1].find_all { |x| x < partition } + [partition]
51      kth_largest_element(lower, k - upper.size)
52    end
53  end
54
55  [
56    { nums: [7, 6, 5, 4, 3, 2, 1], k: 2, x: 6 },
57    { nums: [99, 99], k: 1, x: 99 },
58    { nums: [1], k: 1, x: 1 },
59    { nums: [2, 1], k: 1, x: 2 },
60    { nums: [-1, 2, 0], k: 2, x: 0 },
61    { nums: [-1, 2, 0], k: 3, x: -1 },
62    { nums: [3, 1, 2, 4], k: 2, x: 3 },
63    { nums: [3, 2, 1, 5, 6, 4], k: 2, x: 5 },
64    { nums: [5, 2, 4, 1, 3, 6, 0], k: 2, x: 5 },
65    { nums: [3, 3, 3, 3, 3, 3, 3, 3, 3], k: 8, x: 3 },
66    { nums: [3, 3, 3, 3, 4, 3, 3, 3, 3], k: 1, x: 4 },
67    { nums: [3, 3, 3, 3, 4, 3, 3, 3, 3], k: 5, x: 3 },
68    { nums: [3, 2, 3, 1, 2, 4, 5, 5, 6], k: 4, x: 4 },
69    { nums: [3, 2, 3, 1, 2, 4, 5, 5, 6, 7, 7, 8, 2, 3, 1, 1, 1, 10, 11, 5, 6, 2, 4, 7, 8, 5, 6], k: 1, x: 11 },
70    { nums: [2, 1], k: 2, x: 1 },
71    { nums: [-1, -1], k: 2, x: -1 },
72  ].each do |x|
73    it do
74      expect(kth_largest_element(x[:nums], x[:k])).to eql(x[:x])
75    end
76  end
77
78end