LeetCode 774. Minimize Max Distance to Gas Station


原題鏈接在這里:https://leetcode.com/problems/minimize-max-distance-to-gas-station/description/

題目:

On a horizontal number line, we have gas stations at positions stations[0], stations[1], ..., stations[N-1], where N = stations.length.

Now, we add K more gas stations so that D, the maximum distance between adjacent gas stations, is minimized.

Return the smallest possible value of D.

Example:

Input: stations = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], K = 9
Output: 0.500000

Note:

  1. stations.length will be an integer in range [10, 2000].
  2. stations[i] will be an integer in range [0, 10^8].
  3. K will be an integer in range [1, 10^6].
  4. Answers within 10^-6 of the true value will be accepted as correct.

題解:

在0到10^8范圍內猜測有這么個distance D.

若是符合要求, 那么每兩個station之間的距離除以這個D取整數就是新加油站的數目.

這些新加的數目和若是<= K, it means that distance is big enough. Think like this, if it is able to add j <= K gas stations to get achieve this distance. Than add K-j more just make the distance even smaller. 

所以可以繼續在較小的一側做 binary search to find a smaller D.

Time Complexity: O(nlogW). n = stations.length. W = 10^14. 從10^8范圍猜到10^-6范圍.

Space: O(1).

AC Java:

 1 class Solution {
 2     public double minmaxGasDist(int[] stations, int K) {
 3         double l = 0;
 4         double r = 1e8;
 5         while(r - l > 1e-6){
 6             double m = l + (r-l)/2.0;
 7             if(possible(stations, K, m)){
 8                 r = m;
 9             }else{
10                 l = m;
11             }
12         }
13         
14         return l;
15     }
16     
17     private boolean possible(int [] stations, int K, double d){
18         int count = 0;
19         for(int i = 0; i<stations.length-1; i++){
20             count += (int)((stations[i+1] - stations[i])/d);
21         }
22         
23         return count <= K;
24     }
25 }

類似Koko Eating BananasCapacity To Ship Packages Within D Days.


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