Leetcode 458 - Poor pigs

Note:

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  • First, a pig can drink many buckets at one time.
  • Name all buckets from 0 to n - 1.
  • Start from a simple example, assume there are 1000 buckets, 15 minToTest, 15 minToDie.
  • Because 2^10 > 1000 > 2^9. We need 10 pigs.
  • Allot each pig with a digit of the 10-digit binary number.
  • If a pig dies after drinking all the buckets with digit 1. It means the number of the poisonous bucket has 1 on that digit.
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Question:

There are buckets buckets of liquid, where exactly one of the buckets is poisonous. To figure out which one is poisonous, you feed some number of (poor) pigs the liquid to see whether they will die or not. Unfortunately, you only have minutesToTest minutes to determine which bucket is poisonous.

You can feed the pigs according to these steps:

  1. Choose some live pigs to feed.
  2. For each pig, choose which buckets to feed it. The pig will consume all the chosen buckets simultaneously and will take no time.
  3. Wait for minutesToDie minutes. You may not feed any other pigs during this time.
  4. After minutesToDie minutes have passed, any pigs that have been fed the poisonous bucket will die, and all others will survive.
  5. Repeat this process until you run out of time.

Given buckets, minutesToDie, and minutesToTest, return the minimum number of pigs needed to figure out which bucket is poisonous within the allotted time.

Example:

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Input: buckets = 1000, minutesToDie = 15, minutesToTest = 60
Output: 5

Code:

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/**
* @param {number} buckets
* @param {number} minutesToDie
* @param {number} minutesToTest
* @return {number}
*/
var poorPigs = function(buckets, minutesToDie, minutesToTest) {
const base = ~~(minutesToTest / minutesToDie) + 1;
return Math.ceil(Math.log(buckets) / Math.log(base));
};