Leetcode 2187: Minimum Time to Complete Trips
You are given an array ’time’ where each element ’time[i]’ represents the time taken by the ith bus to complete a trip. Your task is to calculate the minimum amount of time required for all buses to complete at least ’totalTrips’ trips in total. Each bus can complete multiple trips consecutively.
Problem
Approach
Steps
Complexity
Input: The input consists of an array of integers 'time' and an integer 'totalTrips'.
Example: Input: time = [4, 5], totalTrips = 6
Constraints:
• 1 <= time.length <= 10^5
• 1 <= time[i], totalTrips <= 10^7
Output: Return the minimum time required for all buses to complete at least 'totalTrips' trips.
Example: Output: 10
Constraints:
• The solution must handle large inputs efficiently.
Goal: The goal is to find the minimum time required for all buses to complete at least 'totalTrips' trips.
Steps:
• For a given time, calculate the total number of trips completed by all buses.
• Use binary search to find the smallest time for which the total number of trips is at least 'totalTrips'.
Goal: Conditions that the solution must satisfy.
Steps:
• The length of the time array is between 1 and 10^5.
• Each time element is between 1 and 10^7.
• The totalTrips value is between 1 and 10^7.
Assumptions:
• Each bus operates independently and can make multiple trips in succession.
• All buses are in operation at the same time.
• Input: Input: time = [4, 5], totalTrips = 6
• Explanation: At time t = 4, the number of trips completed by the buses are [1,0]. At t = 8, the buses have completed [2,1]. At t = 10, the total number of trips is 6. So, the minimum time required is 10.
• Input: Input: time = [3, 6], totalTrips = 5
• Explanation: At time t = 3, the number of trips completed by the buses are [1,0]. At t = 6, the buses have completed [2,1]. At t = 9, the total number of trips is 5. So, the minimum time required is 9.
Approach: We will calculate the number of trips completed by each bus for a given time and use binary search to find the minimum time to complete the required total trips.
Observations:
• The binary search will allow us to find the optimal time efficiently.
• To solve this problem, we need to determine the number of trips made by all buses for different time intervals and use binary search to narrow down the minimum time.
Steps:
• For each bus, calculate the number of trips completed by the bus in the given time.
• Use binary search to find the smallest time where the total number of trips completed is at least 'totalTrips'.
Empty Inputs:
• Empty input is not possible based on constraints.
Large Inputs:
• Ensure the solution handles cases where time has up to 10^5 elements and the totalTrips can be as large as 10^7.
Special Values:
• If all buses have the same trip time, the solution should still work efficiently.
Constraints:
• The solution must avoid a brute force approach to ensure it works efficiently for large inputs.
long long numberOfTripsForAGivenTime(vector<int> &a, long long int givenTime) {
long long int totalTrips = 0;
for(int &x : a) {
long long int y = x;
totalTrips += givenTime / y;
}
return totalTrips;
}
long long minimumTime(vector<int>& time, int totalTrips) {
long long int low = 1, high = 1e14;
while(low < high) {
long long int mid = low + (high - low) / 2;
if(numberOfTripsForAGivenTime(time, mid) >= totalTrips) {
high = mid;
} else low = mid + 1;
}
return low;
}
1 : Function Definition
long long numberOfTripsForAGivenTime(vector<int> &a, long long int givenTime) {
This defines the function `numberOfTripsForAGivenTime`, which takes a vector of trip durations and a given time, and calculates the total number of trips that can be completed in the given time.
2 : Variable Initialization
long long int totalTrips = 0;
A variable `totalTrips` is initialized to 0 to accumulate the total number of trips that can be made within the given time.
3 : Loop Through Trips
for(int &x : a) {
This loop iterates through each element `x` in the vector `a`, which represents the duration of each trip.
4 : Temporary Variable
long long int y = x;
A temporary variable `y` is created to store the current trip duration `x`. This is done to avoid directly modifying `x` in subsequent calculations.
5 : Accumulate Total Trips
totalTrips += givenTime / y;
The number of trips that can be completed in the given time for the current trip duration `y` is calculated by integer division of `givenTime / y`. This value is added to `totalTrips`.
6 : Return Total Trips
return totalTrips;
The function returns the total number of trips that can be completed within the given time.
7 : Empty Location
This location is left empty as there are no intermediate steps between the functions.
8 : Function Definition
long long minimumTime(vector<int>& time, int totalTrips) {
This defines the function `minimumTime`, which takes a vector of trip durations and the total number of trips to be completed, and calculates the minimum time required to complete those trips.
9 : Binary Search Initialization
long long int low = 1, high = 1e14;
Two variables `low` and `high` are initialized for the binary search. `low` starts at 1 and `high` is set to a very large value (1e14) to cover a wide range of possible times.
10 : Binary Search Loop
while(low < high) {
This loop performs binary search, narrowing down the range of possible times (`low` and `high`) until the correct minimum time is found.
11 : Midpoint Calculation
long long int mid = low + (high - low) / 2;
The midpoint `mid` is calculated as the average of `low` and `high`, which will be checked to see if it allows enough trips to meet or exceed the required `totalTrips`.
12 : Check Trips for Midpoint
if(numberOfTripsForAGivenTime(time, mid) >= totalTrips) {
The function `numberOfTripsForAGivenTime` is called with `mid` as the given time. If the total trips calculated for this time are greater than or equal to the required `totalTrips`, the upper bound (`high`) is adjusted.
13 : Adjust High
high = mid;
If the current `mid` allows enough trips, `high` is updated to `mid` to search for a smaller possible time.
14 : Adjust Low
} else low = mid + 1;
If the current `mid` does not allow enough trips, `low` is updated to `mid + 1` to search for a larger possible time.
15 : Return Minimum Time
return low;
The function returns the minimum time found by the binary search.
Best Case: O(n log T)
Average Case: O(n log T)
Worst Case: O(n log T)
Description: The time complexity is O(n log T), where n is the number of buses and T is the time interval range.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) as we only need a constant amount of extra space for calculations.
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