#include #include #include #include // sorting struct comp { template bool operator()(const T& l, const T& r) const { return (l.second != r.second) ? l.second > r.second : l.first > r.first; } }; // finding the number of occurence of each elements std::map kMostElem(std::vector v) { std::map counter; for (auto & e : v) { auto res = counter.insert(std::pair(e, 1)); if (res.second == false) res.first->second++; } return counter; } int main(void) { std::vector, int>> nums = { { { 1,1,1,2,2,3 }, 2 }, { { 2,2,2,1,1,1 }, 1 }, { { 1 }, 1 } }; std::pair, int> p; for (auto const p : nums) { std::vector num = p.first; int k = p.second; // printing input std::cout << "nums = "; for (size_t i = 0; i < num.size(); i++) std::cout << ((i == 0) ? "[" : "") << num[i] << ((i == num.size() - 1) ? "]," : ","); std::cout << " k = " << k << "\n"; // storing the counter values std::map mTemp = kMostElem(num); // sorting the map in descending order std::set, comp> sortedSet(mTemp.begin(), mTemp.end()); // printing output int temp = 0; for (auto const pair : sortedSet) { if (temp == k) break; std::cout << pair.first << " occurs " << pair.second << " times.\n"; temp++; } std::cout << "\n"; } } /* Given a non-empty array of integers, return the k most frequent elements. Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] Input: nums = [1], k = 1 Output: [1] Note: You may assume k is always valid, 1<=k<=number of unique of elements. Your algorithm's time complexity must be better than O(n log n), where n is the array's size. */

k-most frequent elems