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| 1 | +/* |
| 2 | + * Problem: 973 |
| 3 | + * Name: K Closest Point To Origin |
| 4 | + * Difficulty: Medium |
| 5 | + * Topic: Priority Queue |
| 6 | + * Link: https://leetcode.com/problems/ |
| 7 | + */ |
| 8 | + |
| 9 | +#include <bits/stdc++.h> |
| 10 | +#include <functional> |
| 11 | +#include <utility> |
| 12 | +#include <vector> |
| 13 | +using namespace std; |
| 14 | + |
| 15 | +// Partial Sort Solution |
| 16 | +// Time Complexity: O(n log K) |
| 17 | +// Space Complexity: O(1) |
| 18 | +vector<vector<int>> kClosestPS(vector<vector<int>>& points, int K) { |
| 19 | + partial_sort(points.begin(), points.begin() + K, points.end(), |
| 20 | + [](vector<int>& p, vector<int>& q) { return p[0] * p[0] + p[1] * p[1] < q[0] * q[0] + q[1] * q[1];}); |
| 21 | + return vector<vector<int>>(points.begin(), points.begin() + K); |
| 22 | +} |
| 23 | + |
| 24 | +// Nth_Element Solution |
| 25 | +// Time Complexity: O(n log K) |
| 26 | +// Space Complexity: O(1) |
| 27 | +vector<vector<int>> kClosestNE(vector<vector<int>>& points, int K) { |
| 28 | + nth_element(points.begin(), points.begin() + K - 1, points.end(), |
| 29 | + [](vector<int>& p, vector<int>& q) {return p[0] * p[0] + p[1] * p[1] < q[0] * q[0] + q[1] * q[1];}); |
| 30 | + return vector<vector<int>>(points.begin(), points.begin() + K); |
| 31 | +} |
| 32 | + |
| 33 | +// Min Heap of Points |
| 34 | +// Time Complexity: O(n + k log n) |
| 35 | +// Space Complexity: O(n) |
| 36 | +struct MinDistanceComparator { |
| 37 | + bool operator()(vector<int>& p1, vector<int>& p2){ |
| 38 | + return p1[0] * p1[0] + p1[1] * p1[1] > p2[0] * p2[0] + p2[1] * p2[1]; |
| 39 | + } |
| 40 | +}; |
| 41 | + |
| 42 | +vector<vector<int>> kClosestPQP(vector<vector<int>>& points, int k) { |
| 43 | + priority_queue<vector<int>, vector<vector<int>>, MinDistanceComparator> ordered(points.begin(), points.end()); |
| 44 | + vector<vector<int>> result; |
| 45 | + while (k--){ |
| 46 | + result.push_back(ordered.top()); |
| 47 | + ordered.pop(); |
| 48 | + } |
| 49 | + return result; |
| 50 | +} |
| 51 | + |
| 52 | +// Min Heap of Indexes |
| 53 | +// Time Complexity: O(n + k log n) |
| 54 | +// Space Complexity: O(n) |
| 55 | +vector<vector<int>> kClosestPQI(vector<vector<int>>& points, int k) { |
| 56 | + priority_queue<pair<double, int>, vector<pair<double,int>>, std::greater<pair<double, int>>> ordered; |
| 57 | + for (int idx = 0; idx < points.size(); idx++){ |
| 58 | + ordered.push({points[idx][0] * points[idx][0] + points[idx][1] * points[idx][1], idx}); |
| 59 | + } |
| 60 | + |
| 61 | + vector<vector<int>> result; |
| 62 | + while (k--){ |
| 63 | + result.push_back(points[ordered.top().second]); |
| 64 | + ordered.pop(); |
| 65 | + } |
| 66 | + return result; |
| 67 | +} |
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