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This approach uses binary search to find the minimum time required for the buses to make at least the specified number of trips. The key observation is that if at time `t`, the buses can complete the required trips, then they can also do it in any time greater than `t`. Conversely, if they cannot make the required number of trips at time `t`, then they cannot do it in any lesser time.
We employ binary search with an initial low of 1 and a high value calculated as the maximum possible trip time multiplied by the required total trips. For each midpoint, we calculate the number of trips that all buses can make and adjust the time bounds based on whether the trips meet the requirement or not.
Time Complexity: O(n log(maxTime)), where 'n' is the number of buses and 'maxTime' is the maximum possible time.
Space Complexity: O(1).
1#include <stdio.h>
2
3int canCompleteTrips(int* time, int timeSize, long long t, int totalTrips) {
4 long long
First, we define a helper function canCompleteTrips that verifies if the current time allows the buses to complete the total desired trips. In the main function, we perform binary search on the possible time interval. For each midpoint, we check the fulfillment of the total trips and adjust our search space accordingly.