{"id":8702,"date":"2021-11-14T20:45:28","date_gmt":"2021-11-15T04:45:28","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=8702"},"modified":"2021-11-14T20:49:18","modified_gmt":"2021-11-15T04:49:18","slug":"leetcode-2071-maximum-number-of-tasks-you-can-assign","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/greedy\/leetcode-2071-maximum-number-of-tasks-you-can-assign\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 2071. Maximum Number of Tasks You Can Assign"},"content":{"rendered":"\n

You have n<\/code> tasks and m<\/code> workers. Each task has a strength requirement stored in a 0-indexed<\/strong> integer array tasks<\/code>, with the ith<\/sup><\/code> task requiring tasks[i]<\/code> strength to complete. The strength of each worker is stored in a 0-indexed<\/strong> integer array workers<\/code>, with the jth<\/sup><\/code> worker having workers[j]<\/code> strength. Each worker can only be assigned to a single<\/strong> task and must have a strength greater than or equal<\/strong> to the task’s strength requirement (i.e., workers[j] >= tasks[i]<\/code>).<\/p>\n\n\n\n

Additionally, you have pills<\/code> magical pills that will increase a worker’s strength<\/strong> by strength<\/code>. You can decide which workers receive the magical pills, however, you may only give each worker at most one<\/strong> magical pill.<\/p>\n\n\n\n

Given the 0-indexed <\/strong>integer arrays tasks<\/code> and workers<\/code> and the integers pills<\/code> and strength<\/code>, return the maximum<\/strong> number of tasks that can be completed.<\/em><\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> tasks = [3<\/strong>,2<\/strong>,1<\/strong>], workers = [0<\/strong>,3<\/strong>,3<\/strong>], pills = 1, strength = 1\nOutput:<\/strong> 3\nExplanation:<\/strong>\nWe can assign the magical pill and tasks as follows:\n- Give the magical pill to worker 0.\n- Assign worker 0 to task 2 (0 + 1 >= 1)\n- Assign worker 1 to task 1 (3 >= 2)\n- Assign worker 2 to task 0 (3 >= 3)\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> tasks = [5<\/strong>,4], workers = [0<\/strong>,0,0], pills = 1, strength = 5\nOutput:<\/strong> 1\nExplanation:<\/strong>\nWe can assign the magical pill and tasks as follows:\n- Give the magical pill to worker 0.\n- Assign worker 0 to task 0 (0 + 5 >= 5)\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> tasks = [10<\/strong>,15<\/strong>,30], workers = [0<\/strong>,10<\/strong>,10,10,10], pills = 3, strength = 10\nOutput:<\/strong> 2\nExplanation:<\/strong>\nWe can assign the magical pills and tasks as follows:\n- Give the magical pill to worker 0 and worker 1.\n- Assign worker 0 to task 0 (0 + 10 >= 10)\n- Assign worker 1 to task 1 (10 + 10 >= 15)\n<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> tasks = [5<\/strong>,9,8<\/strong>,5<\/strong>,9], workers = [1,6<\/u><\/strong>,4<\/strong>,2,6<\/strong>], pills = 1, strength = 5\nOutput:<\/strong> 3\nExplanation:<\/strong>\nWe can assign the magical pill and tasks as follows:\n- Give the magical pill to worker 2.\n- Assign worker 1 to task 0 (6 >= 5)\n- Assign worker 2 to task 2 (4 + 5 >= 8)\n- Assign worker 4 to task 3 (6 >= 5)\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • n == tasks.length<\/code><\/li>
  • m == workers.length<\/code><\/li>
  • 1 <= n, m <= 5 * 104<\/sup><\/code><\/li>
  • 0 <= pills <= m<\/code><\/li>
  • 0 <= tasks[i], workers[j], strength <= 109<\/sup><\/code><\/li><\/ul>\n\n\n\n
    Solution: Greedy + Binary Search in Binary Search.<\/strong><\/h2>\n\n\n\n
    Find the smallest k, s.t. we are NOT<\/strong> able to assign. Then answer is k- 1.<\/p>\n\n\n\n
    The key is to verify<\/strong> whether we can assign k tasks or not.<\/p>\n\n\n\n
    Greedy: We want k smallest tasks and k strongest workers.<\/p>\n\n\n\n
    Start with the hardest tasks among (smallest) k:
    1. assign task[i] to the weakest worker without a pill (if he can handle the hardest work so far, then the stronger workers can handle any simpler tasks left)
    2. If 1) is not possible, we find a weakest worker + pill that can handle task[i] (otherwise we are wasting workers)
    3. If 2) is not possible, impossible to finish k tasks.<\/p>\n\n\n\n
    Let k = min(n, m)
    Time complexity: O((logk)2<\/sup> * k)
    Space complexity: O(k)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int maxTaskAssign(vector& tasks, vector& workers, int pills, int strength) {\n    const int n = tasks.size();    \n    sort(begin(tasks), end(tasks));    \n    sort(begin(workers), end(workers)); \n    auto check = [&](int k) {      \n      multiset ws(rbegin(workers), rbegin(workers) + k);\n      for (int i = k - 1, p = 0; i >= 0; --i) {\n        auto it = ws.lower_bound(tasks[i]);\n        if (it != end(ws)) {\n          ws.erase(it);\n        } else if ((it = ws.lower_bound(tasks[i] - strength)) != end(ws)) {\n          if (++p > pills) return false;\n          ws.erase(it);\n        } else {\n          return false;\n        }\n      }\n      return true;\n    };\n    int l = 0;\n    int r = min(tasks.size(), workers.size()) + 1;\n    while (l < r) {\n      int m = l + (r - l) \/ 2;\n      if (!check(m)) {\n        r = m;\n      } else {\n        l = m + 1;\n      }\n    }\n    return l - 1;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
    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…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[149,51],"tags":[52,217],"class_list":["post-8702","post","type-post","status-publish","format-standard","hentry","category-binary-search","category-greedy","tag-binary-search","tag-hard","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8702","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=8702"}],"version-history":[{"count":3,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8702\/revisions"}],"predecessor-version":[{"id":8706,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8702\/revisions\/8706"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=8702"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=8702"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=8702"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}