You are given a 0-indexed integer array costs where costs[i] is the cost of hiring the ith worker.
You are also given two integers k and candidates. We want to hire exactly k workers according to the following rules:
k sessions and hire exactly one worker in each session.candidates workers or the last candidates workers. Break the tie by the smallest index.
costs = [3,2,7,7,1,2] and candidates = 2, then in the first hiring session, we will choose the 4th worker because they have the lowest cost [3,2,7,7,1,2].1st worker because they have the same lowest cost as 4th worker but they have the smallest index [3,2,7,7,2]. Please note that the indexing may be changed in the process.Return the total cost to hire exactly k workers.
Example 1:
Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4 Output: 11 Explanation: We hire 3 workers in total. The total cost is initially 0. - In the first hiring round we choose the worker from [17,12,10,2,7,2,11,20,8]. The lowest cost is 2, and we break the tie by the smallest index, which is 3. The total cost = 0 + 2 = 2. - In the second hiring round we choose the worker from [17,12,10,7,2,11,20,8]. The lowest cost is 2 (index 4). The total cost = 2 + 2 = 4. - In the third hiring round we choose the worker from [17,12,10,7,11,20,8]. The lowest cost is 7 (index 3). The total cost = 4 + 7 = 11. Notice that the worker with index 3 was common in the first and last four workers. The total hiring cost is 11.
Example 2:
Input: costs = [1,2,4,1], k = 3, candidates = 3 Output: 4 Explanation: We hire 3 workers in total. The total cost is initially 0. - In the first hiring round we choose the worker from [1,2,4,1]. The lowest cost is 1, and we break the tie by the smallest index, which is 0. The total cost = 0 + 1 = 1. Notice that workers with index 1 and 2 are common in the first and last 3 workers. - In the second hiring round we choose the worker from [2,4,1]. The lowest cost is 1 (index 2). The total cost = 1 + 1 = 2. - In the third hiring round there are less than three candidates. We choose the worker from the remaining workers [2,4]. The lowest cost is 2 (index 0). The total cost = 2 + 2 = 4. The total hiring cost is 4.
Constraints:
1 <= costs.length <= 105 1 <= costs[i] <= 1051 <= k, candidates <= costs.lengthclass Solution {
public:
struct cmp{
bool operator()(pair<int, bool> a, pair<int, bool> b){
if(a.first==b.first)
return b.second;
return a.first>b.first;
}
};
long long totalCost(vector<int>& costs, int k, int candidates) {
long long int res=0;
priority_queue <pair<int, bool>, vector<pair<int, bool>>, cmp> pq; // {cost, fromLt}
bool complete = false;
int lt=0, rt=costs.size()-1;
for(; lt<candidates; lt++)
pq.push({costs[lt], true});
for(; rt>lt && rt>costs.size()-1-candidates; rt--)
pq.push({costs[rt], false});
if(rt<lt) complete = true;
while(k--){
auto curr = pq.top();
pq.pop();
res+= curr.first;
if(complete)
continue;
curr.second? pq.push({costs[lt++], true}): pq.push({costs[rt--], false});
if(rt<lt) complete = true;
}
return res;
}
};