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<<-DOC
Find the kth largest element in an unsorted array. This will be the kth largest element in sorted order, not the kth distinct element.
Example
For nums = [7, 6, 5, 4, 3, 2, 1] and k = 2, the output should be
kthLargestElement(nums, k) = 6;
For nums = [99, 99] and k = 1, the output should be
kthLargestElement(nums, k) = 99.
Input/Output
[time limit] 4000ms (rb)
[input] array.integer nums
Guaranteed constraints:
1 ≤ nums.length ≤ 105,
-105 ≤ nums[i] ≤ 105.
[input] integer k
Guaranteed constraints:
1 ≤ k ≤ nums.length.
[output] integer
DOC
describe "#kth_largest_element" do
def kth_largest_element(numbers, k)
numbers.sort[-k]
end
#def kth_largest_element(numbers, k)
#items = Array.new(105)
#numbers.each do |n|
#items[n] = n
#end
#items.compact[-k]
#end
def kth_largest_element(numbers, k)
return numbers[0] if numbers.size == 1
numbers.shuffle!
partition = numbers[0]
upper = numbers[1..-1].find_all { |x| x >= partition }
if upper.size >= k
kth_largest_element(upper, k)
else
lower = numbers[1..-1].find_all { |x| x < partition } + [partition]
kth_largest_element(lower, k - upper.size)
end
end
[
{ nums: [7, 6, 5, 4, 3, 2, 1], k: 2, x: 6 },
{ nums: [99, 99], k: 1, x: 99 },
{ nums: [1], k: 1, x: 1 },
{ nums: [2, 1], k: 1, x: 2 },
{ nums: [-1, 2, 0], k: 2, x: 0 },
{ nums: [-1, 2, 0], k: 3, x: -1 },
{ nums: [3, 1, 2, 4], k: 2, x: 3 },
{ nums: [3, 2, 1, 5, 6, 4], k: 2, x: 5 },
{ nums: [5, 2, 4, 1, 3, 6, 0], k: 2, x: 5 },
{ nums: [3, 3, 3, 3, 3, 3, 3, 3, 3], k: 8, x: 3 },
{ nums: [3, 3, 3, 3, 4, 3, 3, 3, 3], k: 1, x: 4 },
{ nums: [3, 3, 3, 3, 4, 3, 3, 3, 3], k: 5, x: 3 },
{ nums: [3, 2, 3, 1, 2, 4, 5, 5, 6], k: 4, x: 4 },
{ 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 },
{ nums: [2, 1], k: 2, x: 1 },
{ nums: [-1, -1], k: 2, x: -1 },
].each do |x|
it do
expect(kth_largest_element(x[:nums], x[:k])).to eql(x[:x])
end
end
end
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