Minimum Time to Complete Trips

• Initialize a variable currentTime to 1.
• Start an infinite loop.
• Inside the loop, initialize tripsCompleted to 0.
• Iterate through each busTime in the time array.
• For each busTime, calculate the trips completed by currentTime as currentTime / busTime and add it to tripsCompleted.
• After checking all buses, if tripsCompleted is greater than or equal to totalTrips, then currentTime is the minimum time. Return currentTime.
• If not, increment currentTime and continue the loop.

The algorithm iterates through time, starting from t = 1. In each iteration, it calculates the total number of trips possible by that time.

To calculate the total trips for a given time t, we iterate through the time array. For each bus i with trip time time[i], the number of trips it can complete is t / time[i]. We sum these values for all buses.

If the calculated total trips is at least totalTrips, we have found our answer, and we return the current time t.

This process continues until the condition is met. However, the maximum possible answer can be very large (up to 10^14), making this approach too slow for the given constraints.

1class Solution {2    public long minimumTime(int[] time, int totalTrips) {3        // This approach is too slow and will time out.4        long currentTime = 1;5        while (true) {6            long tripsCompleted = 0;7            for (int t : time) {8                // Using long for currentTime to avoid overflow issues, though9                // this loop will time out long before overflow becomes a problem.10                tripsCompleted += currentTime / t;11            }12            if (tripsCompleted >= totalTrips) {13                return currentTime;14            }15            currentTime++;16        }17    }18}

• Determine the search range for the time. The lower bound low is 1. A good upper bound high is the time taken by the fastest bus to complete all trips alone, i.e., min(time) * totalTrips. Use long for these variables to prevent overflow.
• Initialize a variable ans to high to store the minimum time found so far.
• While low <= high:
  • Calculate the midpoint mid = low + (high - low) / 2.
  • Check if it's possible to complete totalTrips within mid time. This is done by calculating trips = sum(mid / t) for each t in time.
  • If trips >= totalTrips:
    • mid is a valid time. It might be the minimum, so we store it: ans = mid.
    • Try to find an even smaller time by searching in the left half: high = mid - 1.
  • Else (trips < totalTrips):
    • mid is not enough time. We need more time, so search in the right half: low = mid + 1.
• After the loop terminates, ans will hold the minimum time required.

We first define a search space for the possible answer (time). The lower bound can be 1. A reasonable upper bound is the time it would take for the fastest bus to complete all the trips by itself, which is min(time) * totalTrips. Since this value can be very large, we must use long to avoid integer overflow.

We then perform a binary search within this range [low, high].

In each step of the binary search, we take the middle time mid and check if it's possible to complete totalTrips within this time. We do this by creating a helper function canComplete(mid) which calculates the total trips sum(mid / time[i]) and checks if it's >= totalTrips.

• If canComplete(mid) is true, it means mid is a potential answer, and there might be an even smaller time that works. So, we record mid as a possible answer and shrink our search space to the left half: high = mid - 1.
• If canComplete(mid) is false, mid is too small. We need more time, so we search in the right half: low = mid + 1.

The search continues until low exceeds high. The last valid time we found is the minimum time.

1class Solution {2    public long minimumTime(int[] time, int totalTrips) {3        long low = 1;4        long minTime = Long.MAX_VALUE;5        for (int t : time) {6            minTime = Math.min(minTime, t);7        }8        long high = minTime * totalTrips; // A reasonable upper bound9        long ans = high;10 11        while (low <= high) {12            long mid = low + (high - low) / 2;13            if (isPossible(time, totalTrips, mid)) {14                ans = mid;15                high = mid - 1;16            } else {17                low = mid + 1;18            }19        }20        return ans;21    }22 23    private boolean isPossible(int[] time, int totalTrips, long givenTime) {24        long tripsCompleted = 0;25        for (int t : time) {26            tripsCompleted += givenTime / t;27            // An early exit optimization: if we've already met the goal,28            // no need to calculate further.29            if (tripsCompleted >= totalTrips) {30                return true;31            }32        }33        return false;34    }35}