subset_min_max_count.cpp

puzzles/interviews/training/subset_min_max_count.cpp

@@ -0,0 +1,74 @@
+/*
+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;
+}}