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-rw-r--r--other/max_num_tasks_assign.cpp68
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diff --git a/other/max_num_tasks_assign.cpp b/other/max_num_tasks_assign.cpp
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-// 2071. Maximum Number of Tasks You Can Assign
-#include "leetcode.h"
-
-/*
- * you have 'n' tasks and 'm' workers. each task has a strength requirement
- * stored in a 0-indexed integer array 'tasks', with the i'th task requiring
- * 'tasks[i]' strength to complete. the strength of each worker is stored in a
- * 0-indexed array 'workers' with the j'th worker having 'workers[j]' strength.
- * each worker can only be assigned to a single task and must have a strength
- * greather than or equal to the task's strength requirement. additionally, you
- * have 'pills' magic pills that will increase a workers strength by 'strength'.
- * you can decide which workers receive the 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.
- */
-
-class Solution {
-public:
- int maxTaskAssign(vector<int> &tasks, vector<int> &workers, int pills,
- int strength) {
- int n = tasks.size(), m = workers.size();
- sort(tasks.begin(), tasks.end());
- sort(workers.begin(), workers.end());
- int lo = 0, hi = min(m, n), ans;
- while (lo <= hi) {
- int mid = lo + (hi - lo) / 2;
- int cnt = 0;
- bool flag = true;
- multiset<int> st(workers.begin(), workers.end());
- for (int i = mid - 1; i >= 0; i--) {
- auto it = prev(st.end());
- if (tasks[i] <= *it)
- st.erase(it);
- else {
- auto it = st.lower_bound(tasks[i] - strength);
- if (it != st.end()) {
- cnt++;
- st.erase(it);
- } else {
- flag = false;
- break;
- }
- }
- if (cnt > pills) {
- flag = false;
- break;
- }
- }
- if (flag) {
- ans = mid;
- lo = mid + 1;
- } else {
- hi = mid - 1;
- }
- }
- return ans;
- }
-};
-
-int main() {
- Solution obj;
- vector<int> t1 = {3, 2, 1}, t2 = {5, 4}, t3 = {10, 15, 30};
- vector<int> w1 = {0, 3, 3}, w2 = {0, 0, 0}, w3 = {0, 10, 10, 10, 10};
- printf("%d\n", obj.maxTaskAssign(t1, w1, 1, 1)); // expect: 3
- printf("%d\n", obj.maxTaskAssign(t2, w2, 1, 5)); // expect: 1
- printf("%d\n", obj.maxTaskAssign(t3, w3, 3, 10)); // expect: 2
-}