Find the maximum cost of an array of pairs choosing at most K pairs
Given an array of pairs arr[], the task is to find the maximum cost choosing at most K pairs. The cost of an array of pairs is defined as product of the sum of first elements of the selected pair and the minimum among the second elements of the selected pairs. For example, if the following pairs are selected (3, 7), (9, 2) and (2, 5), then the cost will be (3+9+2)*(2) = 28.
Examples:
Input: arr[] = { {4, 7}, {15, 1}, {3, 6}, {6, 8} }, K = 3
Output: 78
The pairs 1, 3 and 4 are selected, therefore the cost = (4 + 3 + 6) * 6 = 78.Input: arr[] = { {62, 21}, {31, 16}, {19, 2}, {32, 19}, {12, 17} }, K = 4
Output: 2192
Approach: If the second element of a pair is fixed in the answer, then K-1(or less) other pairs are to be selected from those pairs whose second element is greater or equal to the fixed second element and the answer will be maximum if those are chosen such that the sum of first elements is maximum. So, sort the array according to second element and then iterate in descending order taking maximum sum of the first element of K pairs(or less). The maximum sum of first element K pairs can be taken with the help of Set data structure.
Below is the implementation of the above approach:
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Driver function to sort the array elements// by second element of pairsbool sortbysec(const pair<int, int>& a, const pair<int, int>& b){ return (a.second < b.second);} // Function that returns the maximum cost of// an array of pairs choosing at most K pairs.int maxCost(pair<int, int> a[], int N, int K){ // Initialize result and temporary sum variables int res = 0, sum = 0; // Initialize Set to store K greatest // element for maximum sum set<pair<int, int> > s; // Sort array by second element sort(a, a + N, sortbysec); // Iterate in descending order for (int i = N - 1; i >= 0; --i) { s.insert(make_pair(a[i].first, i)); sum += a[i].first; while (s.size() > K) { auto it = s.begin(); sum -= it->first; s.erase(it); } res = max(res, sum * a[i].second); } return res;} // Driver Codeint main(){ pair<int, int> arr[] = { { 12, 3 }, { 62, 21 }, { 31, 16 }, { 19, 2 }, { 32, 19 }, { 12, 17 }, { 1, 7 } }; int N = sizeof(arr) / sizeof(arr[0]); int K = 3; cout << maxCost(arr, N, K); return 0;} |
2000
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