{"id":9238,"date":"2021-12-25T18:41:45","date_gmt":"2021-12-26T02:41:45","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=9238"},"modified":"2021-12-25T18:42:59","modified_gmt":"2021-12-26T02:42:59","slug":"leetcode-1921-eliminate-maximum-number-of-monsters","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/greedy\/leetcode-1921-eliminate-maximum-number-of-monsters\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1921. Eliminate Maximum Number of Monsters"},"content":{"rendered":"\n

You are playing a video game where you are defending your city from a group of n<\/code> monsters. You are given a 0-indexed<\/strong> integer array dist<\/code> of size n<\/code>, where dist[i]<\/code> is the initial distance<\/strong> in kilometers of the ith<\/sup><\/code> monster from the city.<\/p>\n\n\n\n

The monsters walk toward the city at a constant<\/strong> speed. The speed of each monster is given to you in an integer array speed<\/code> of size n<\/code>, where speed[i]<\/code> is the speed of the ith<\/sup><\/code> monster in kilometers per minute.<\/p>\n\n\n\n

You have a weapon that, once fully charged, can eliminate a single<\/strong> monster. However, the weapon takes one minute<\/strong> to charge.The weapon is fully charged at the very start.<\/p>\n\n\n\n

You lose when any monster reaches your city. If a monster reaches the city at the exact moment the weapon is fully charged, it counts as a loss<\/strong>, and the game ends before you can use your weapon.<\/p>\n\n\n\n

Return the maximum<\/strong> number of monsters that you can eliminate before you lose, or <\/em>n<\/code> if you can eliminate all the monsters before they reach the city.<\/em><\/p>\n\n\n\n

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

Input:<\/strong> dist = [1,3,4], speed = [1,1,1]\nOutput:<\/strong> 3\nExplanation:<\/strong>\nIn the beginning, the distances of the monsters are [1,3,4]. You eliminate the first monster.\nAfter a minute, the distances of the monsters are [X,2,3]. You eliminate the second monster.\nAfter a minute, the distances of the monsters are [X,X,2]. You eliminate the thrid monster.\nAll 3 monsters can be eliminated.<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> dist = [1,1,2,3], speed = [1,1,1,1]\nOutput:<\/strong> 1\nExplanation:<\/strong>\nIn the beginning, the distances of the monsters are [1,1,2,3]. You eliminate the first monster.\nAfter a minute, the distances of the monsters are [X,0,1,2], so you lose.\nYou can only eliminate 1 monster.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> dist = [3,2,4], speed = [5,3,2]\nOutput:<\/strong> 1\nExplanation:<\/strong>\nIn the beginning, the distances of the monsters are [3,2,4]. You eliminate the first monster.\nAfter a minute, the distances of the monsters are [X,0,2], so you lose.\nYou can only eliminate 1 monster.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • n == dist.length == speed.length<\/code><\/li>
  • 1 <= n <= 105<\/sup><\/code><\/li>
  • 1 <= dist[i], speed[i] <= 105<\/sup><\/code><\/li><\/ul>\n\n\n\n
    Solution: Greedy<\/strong><\/h2>\n\n\n\n
    Sort by arrival time, and see how many we can eliminate.<\/p>\n\n\n\n
    Time complexity: O(nlogn)
    Space complexity: O(n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int eliminateMaximum(vector& dist, vector& speed) {\n    const int n = dist.size();\n    vector t(n);\n    for (int i = 0; i < n; ++i)\n      t[i] = (dist[i] + speed[i] - 1) \/ speed[i];\n    sort(begin(t), end(t));    \n    for (int i = 0; i < n; ++i)\n      if (t[i] <= i) return i;\n    return n;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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