75 lines
1.7 KiB
C++
75 lines
1.7 KiB
C++
/*
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VIM: let b:lcppflags="-std=c++14 -O2 -pthread -I."
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VIM: let b:wcppflags="/O2 /EHsc /DWIN32"
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*/
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#include "stdafx.h"
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#include <iostream>
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#include <exception>
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#include <chrono>
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#include <vector>
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#include <algorithm>
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/*
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google-interview-questions
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For a given vector of integers and integer K, find the number of non-empty subsets S such that min(S) + max(S) <= K
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For example, for K = 8 and vector [2, 4, 5, 7], the solution is 5 ([2], [4], [2, 4], [2, 4, 5], [2, 5])
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The time complexity should be O(n2). Approach and code was asked
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*/
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int countSubsets(std::vector<int> numbers, int k)
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{
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int total_sum = 0;
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for (int i = 0; i < numbers.size(); ++i) {
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if (2 * numbers[i] > k) {
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continue;
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}
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int sum = 1;
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int max = numbers[i];
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int min = numbers[i];
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for (int j = i+1; j < numbers.size(); ++j) {
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int tmp_max = std::max(max, numbers[j]);
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int tmp_min = std::min(min, numbers[j]);
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if (tmp_max + tmp_min <= k) {
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sum *= 2;
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min = tmp_min;
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max = tmp_max;
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}
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}
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total_sum += sum;
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}
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return total_sum;
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}
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int main()
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{try{
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auto begin = std::chrono::high_resolution_clock::now();
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//......
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std::cout << countSubsets({ 2, 4, 5, 7 }, 8) << std::endl;
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auto end = std::chrono::high_resolution_clock::now();
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std::chrono::duration<double> seconds = end - begin;
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std::cout << "Time: " << seconds.count() << std::endl;
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return 0;
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}
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catch ( const std::exception& e )
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{
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std::cerr << std::endl
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<< "std::exception(\"" << e.what() << "\")." << std::endl;
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return 2;
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}
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catch ( ... )
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{
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std::cerr << std::endl
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<< "unknown exception." << std::endl;
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return 1;
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}}
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