Description Link to heading
- Maximum Number of Tasks You Can Assign (Hard)
You have n
tasks and m
workers. Each task has a strength requirement stored in a 0-indexed
integer array tasks
, with the ith
task requiring tasks[i]
strength to complete. The strength
of each worker is stored in a 0-indexed integer array workers
, with the jth
worker having
workers[j]
strength. Each worker can only be assigned to a single task and must have a
strength greater than or equal to the task’s strength requirement (i.e., workers[j] >= tasks[i]
).
Additionally, you have pills
magical pills that will increase a worker’s strength by
strength
. You can decide which workers receive the magical pills, however, you may only give each
worker at most one magical pill.
Given the 0-indexed integer arrays tasks
and workers
and the integers pills
and
strength
, return the maximum number of tasks that can be completed.
Example 1:
Input: tasks = [3,2,1], workers = [0,3,3], pills = 1, strength = 1
Output: 3
Explanation:
We can assign the magical pill and tasks as follows:
- Give the magical pill to worker 0.
- Assign worker 0 to task 2 (0 + 1 >= 1)
- Assign worker 1 to task 1 (3 >= 2)
- Assign worker 2 to task 0 (3 >= 3)
Example 2:
Input: tasks = [5,4], workers = [0,0,0], pills = 1, strength = 5
Output: 1
Explanation:
We can assign the magical pill and tasks as follows:
- Give the magical pill to worker 0.
- Assign worker 0 to task 0 (0 + 5 >= 5)
Example 3:
Input: tasks = [10,15,30], workers = [0,10,10,10,10], pills = 3, strength = 10
Output: 2
Explanation:
We can assign the magical pills and tasks as follows:
- Give the magical pill to worker 0 and worker 1.
- Assign worker 0 to task 0 (0 + 10 >= 10)
- Assign worker 1 to task 1 (10 + 10 >= 15)
The last pill is not given because it will not make any worker strong enough for the last task.
Constraints:
n == tasks.length
m == workers.length
1 <= n, m <= 5 * 10⁴
0 <= pills <= m
0 <= tasks[i], workers[j], strength <= 10⁹
Solution Link to heading
Binary search the answer.
Code Link to heading
class Solution {
public:
bool check(int num, vector<int> &tasks, vector<int> &workers, int pills, map<int, int> mp, int strength) {
for (int i = 0, j = workers.size() - num; i < tasks.size() && j < workers.size(); ++i, ++j) {
if (mp.begin()->first > workers[j]) {
if (pills <= 0) {
return false;
} else {
if (mp.begin()->first > strength + workers[j]) {
return false;
} else {
auto iter = mp.lower_bound(strength + workers[j]);
if (iter->first == strength + workers[j]) {
--iter->second;
if (iter->second == 0) {
mp.erase(iter);
}
} else {
--iter;
--iter->second;
if (iter->second == 0) {
mp.erase(iter);
}
}
--pills;
}
}
} else {
mp.begin()->second--;
if (mp.begin()->second == 0) {
mp.erase(mp.begin());
}
}
}
return true;
}
int maxTaskAssign(vector<int> &tasks, vector<int> &workers, int pills, int strength) {
int m = std::min(workers.size(), tasks.size());
std::sort(tasks.begin(), tasks.end());
std::sort(workers.begin(), workers.end());
map<int, int> mp;
for (int i = 0; i < tasks.size(); i++) {
mp[tasks[i]]++;
}
int left = 0, right = m;
while (left < right) {
int mid = left + (right - left) / 2;
if (check(mid, tasks, workers, pills, mp, strength)) {
left = mid + 1;
} else {
right = mid;
}
}
if (check(left, tasks, workers, pills, mp, strength))
return left;
return left - 1;
}
};