/*
VIM: let b:lcppflags="-std=c++14 -O2 -pthread -I."
VIM: let b:wcppflags="/O2 /EHsc /DWIN32"
*/
#include "stdafx.h"
#include <iostream>
#include <exception>
#include <chrono>
#include <vector>
#include <algorithm>

/*
 google-interview-questions
    For a given vector of integers and integer K, find the number of non-empty subsets S such that min(S) + max(S) <= K
    For example, for K = 8 and vector [2, 4, 5, 7], the solution is 5 ([2], [4], [2, 4], [2, 4, 5], [2, 5])
    The time complexity should be O(n2). Approach and code was asked
*/

int countSubsets(std::vector<int> numbers, int k)
{
	int total_sum = 0;
	for (int i = 0; i < numbers.size(); ++i) {

		if (2 * numbers[i] > k) {
			continue;
		}

		int sum = 1;
		int max = numbers[i];
		int min = numbers[i];
		for (int j = i+1; j < numbers.size(); ++j) {
			int tmp_max = std::max(max, numbers[j]);
			int tmp_min = std::min(min, numbers[j]);

			if (tmp_max + tmp_min <= k) {
				sum *= 2;
				min = tmp_min;
				max = tmp_max;
			}
		}
		total_sum += sum;
	}

	return total_sum;
}

int main()
{try{
    auto begin = std::chrono::high_resolution_clock::now();


    //......
	std::cout << countSubsets({ 2, 4, 5, 7 }, 8) << std::endl;




    auto end = std::chrono::high_resolution_clock::now();
    std::chrono::duration<double> seconds = end - begin;
    std::cout << "Time: " << seconds.count() << std::endl;
    return 0;
}
catch ( const std::exception& e )
{
    std::cerr << std::endl
            << "std::exception(\"" << e.what() << "\")." << std::endl;
    return 2;
}
catch ( ... )
{
    std::cerr  << std::endl
            << "unknown exception." << std::endl;
    return 1;
}}