{"id":7028,"date":"2020-07-05T08:48:27","date_gmt":"2020-07-05T15:48:27","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=7028"},"modified":"2020-07-08T17:18:44","modified_gmt":"2020-07-09T00:18:44","slug":"leetcode-1503-last-moment-before-all-ants-fall-out-of-a-plank","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/math\/leetcode-1503-last-moment-before-all-ants-fall-out-of-a-plank\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1503. Last Moment Before All Ants Fall Out of a Plank"},"content":{"rendered":"\n

We have a wooden plank of the length n<\/code> units<\/strong>. Some ants are walking on the plank, each ant moves with speed 1 unit per second<\/strong>. Some of the ants move to the left<\/strong>, the other move to the right<\/strong>.<\/p>\n\n\n\n

When two ants moving in two different<\/strong> directions meet at some point, they change their directions and continue moving again. Assume changing directions doesn’t take any additional time.<\/p>\n\n\n\n

When an ant reaches one end<\/strong> of the plank at a time t<\/code>, it falls out of the plank imediately.<\/p>\n\n\n\n

Given an integer n<\/code> and two integer arrays left<\/code> and right<\/code>, the positions of the ants moving to the left and the right. Return the moment<\/em> when the last ant(s) fall out of the plank.<\/p>\n\n\n\n

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

\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> n = 4, left = [4,3], right = [0,1]\nOutput:<\/strong> 4\nExplanation:<\/strong> In the image above:\n-The ant at index 0 is named A and going to the right.\n-The ant at index 1 is named B and going to the right.\n-The ant at index 3 is named C and going to the left.\n-The ant at index 4 is named D and going to the left.\nNote that the last moment when an ant was on the plank is t = 4 second, after that it falls imediately out of the plank. (i.e. We can say that at t = 4.0000000001, there is no ants on the plank).\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> n = 7, left = [], right = [0,1,2,3,4,5,6,7]\nOutput:<\/strong> 7\nExplanation:<\/strong> All ants are going to the right, the ant at index 0 needs 7 seconds to fall.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> n = 7, left = [0,1,2,3,4,5,6,7], right = []\nOutput:<\/strong> 7\nExplanation:<\/strong> All ants are going to the left, the ant at index 7 needs 7 seconds to fall.\n<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> n = 9, left = [5], right = [4]\nOutput:<\/strong> 5\nExplanation:<\/strong> At t = 1 second, both ants will be at the same intial position but with different direction.\n<\/pre>\n\n\n\n
Example 5:<\/strong><\/p>\n\n\n\n
Input:<\/strong> n = 6, left = [6], right = [0]\nOutput:<\/strong> 6\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= n <= 10^4<\/code><\/li>
  • 0 <= left.length <= n + 1<\/code><\/li>
  • 0 <= left[i] <= n<\/code><\/li>
  • 0 <= right.length <= n + 1<\/code><\/li>
  • 0 <= right[i] <= n<\/code><\/li>
  • 1 <= left.length + right.length <= n + 1<\/code><\/li>
  • All values of left<\/code> and right<\/code> are unique, and each value can appear only in one<\/strong> of the two arrays.<\/li><\/ul>\n\n\n\n
    Solution: Keep Walking<\/strong><\/h2>\n\n\n\n
    When two ants A –> and <– B meet at some point, they change directions <– A  B –>, we can swap the ids of the ants as <– B A–>, so it’s the same as walking individually and passed by. Then we just need to find the max\/min of the left\/right arrays.<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int getLastMoment(int n, vector& left, vector& right) {\n    const int t1 = left.empty() ? 0 : *max_element(cbegin(left), cend(left));\n    const int t2 = right.empty() ? 0 : n - *min_element(cbegin(right), cend(right));\n    return max(t1, t2);\n  }\n};\n<\/pre>\n\n<\/div>
    Java<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\n  public int getLastMoment(int n, int[] left, int[] right) {\n    int t1 = left.length == 0 ? 0 : Arrays.stream(left).max().getAsInt();\n    int t2 = right.length == 0 ? 0 : n - Arrays.stream(right).min().getAsInt();\n    return Math.max(t1, t2);\n  }\n}\n<\/pre>\n\n<\/div>
    Python3<\/h2>\n
    \n\n
    \n# Author: Huahua\nclass Solution:\n  def getLastMoment(self, n: int, left: List[int], right: List[int]) -> int:\n    t1 = max(left) if left else 0\n    t2 = n - min(right) if right else 0\n    return max(t1, t2)\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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