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
Example 1:
Input：stations = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]，K = 9
Output：0.50
Explanation：The distance between adjacent gas stations is 0.50
Example 2:
Input：stations = [3,6,12,19,33,44,67,72,89,95]，K = 2
Output：14.00
Explanation：construction of gas stations at 86 locations
Notice
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.
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.
Notes:
This question can be solved by the binary search method.
Apparently, the question is asking for a min of maximum. Let say minMax, when val >= minMax, we can always finish one event; otherwise, it cannot be done. And val is apparently in a range.
So it is a binary search!
See the code below:
class Solution {
public:
/**
* @param stations: an integer array
* @param k: an integer
* @return: the smallest possible value of D
*/
double minmaxGasDist(vector<int> &stations, int k) {
// Write your code here
sort(stations.begin(), stations.end());
double epslon = 0.00001, left = epslon, right = stations.back();
while(left + epslon < right) {
double mid = left + (right - left) / 2;
if(!isValid(stations, k, mid)) left = mid;
else right = mid;
}
return left;
}
private:
bool isValid(vector<int>& gs, int k, double mid) {
int ct = 0;
double sum = 0;
for(int i=0; i+1<gs.size(); ++i) {
double dis = (double) (gs[i+1] - gs[i]);
ct += (int)(dis/mid);
}
return ct <= k;
}
};
