LeetCode 1883: Minimum Skips to Arrive at Meeting On Time Solution
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1883. Minimum Skips to Arrive at Meeting On Time
Problem Explanation
Explanation
To solve this problem, we can use binary search to find the minimum number of skips required to arrive at the meeting on time. We can define the search space as the possible number of skips and then check if it is possible to arrive at the meeting on time with that number of skips. If it is possible, we can try to minimize the skips by moving the right boundary of the search space. If it is not possible, we need to increase the number of skips by moving the left boundary of the search space.
We can simulate the traversal of roads while considering the rest times and calculate the total time taken to reach the meeting. If the total time is less than or equal to the given hours before the meeting, we update the answer with the current number of skips.
The time complexity of this approach is O(n * log(maxSkips)), where n is the number of roads and maxSkips is the maximum number of skips needed to consider. The space complexity is O(1).
Solution Code
class Solution {
public int minSkips(int[] dist, int speed, int hoursBefore) {
int n = dist.length;
int left = 0, right = 10000000;
while (left < right) {
int skips = left + (right - left) / 2;
double time = 0;
boolean possible = true;
for (int i = 0; i < n; i++) {
time = Math.ceil(time);
time += (double) dist[i] / speed;
if (time > hoursBefore) {
possible = false;
break;
}
}
if (possible) {
right = skips;
} else {
left = skips + 1;
}
}
return left == 10000000 ? -1 : left;
}
}Try It Yourself
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