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Find the maximum cost of an array of pairs choosing at most K pairs in C++
Suppose we have an array of pairs A; we have to find the maximum cost for selecting at most K pairs. In this case, the cost of an array of pairs type elements is the product of the sum of first elements of the selected pair and the smallest among the second elements of the selected pairs. As an example, if these pairs are selected (4, 8), (10, 3) and (3, 6), then the cost will be (4+10+3)*(3) = 51, for K=3
So, if the input is like A = [(15, 5), (65, 25), (35, 20), (20, 5), (35, 20), (15, 18), (3, 8), (12, 17)], K = 4, then the output will be 2700
To solve this, we will follow these steps −
res := 0, sum = 0
N := size of A
Define one set of pairs called my_set
sort the array A based on the second value of each pair
for initialize i := N - 1, when i >= 0, update (decrease i by 1), do −
make a pair (first element of A[i], i) and insert into my_set
sum := sum + first element of A[i]
while size of my_set > K, do −
it := first element of my_set
sum := sum - first of it
delete it from my_set
res := maximum of res and sum * second of A[i]
return res
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
bool compactor(const pair<int, int>& a,const pair<int, int>& b) {
return (a.second < b.second);
}
int get_maximum_cost(vector<pair<int, int> > &A, int K){
int res = 0, sum = 0;
int N = A.size();
set<pair<int, int>> my_set;
sort(A.begin(), A.end(), compactor);
for (int i = N - 1; i >= 0; --i) {
my_set.insert(make_pair(A[i].first, i));
sum += A[i].first;
while (my_set.size() > K) {
auto it = my_set.begin();
sum -= it->first;
my_set.erase(it);
}
res = max(res, sum * A[i].second);
}
return res;
}
int main() {
vector<pair<int, int> > arr = {{ 15, 5}, { 65, 25}, { 35, 20}, { 20, 5}, { 35, 20}, { 15, 18}, { 3, 8 }, {12, 17}};
int K = 4;
cout << get_maximum_cost(arr, K);
}Input
{{15, 5},{65, 25}, { 35, 20}, { 20, 5}, { 35, 20}, { 15, 18}, { 3, 8
}, {12, 17}}, 4Output
2700
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