1. Minimum Total Weight

There are chocolates, and the weight of the chocolates is given as an array of integers arranged. We are required to eat the weight of the dth chocolate every day; one can pick one remaining chocolate, d days later. The total weight of the chocolates after d days is the sum of all the remaining chocolates after d days. Make sure that on each day, the same chocolate is not picked twice. The weight of the first eaten can be calculated as floor(W/d), where W is the minimum remaining weight.

Example

Chocolates = [30, 25, 12]

d = 4

Day	Weight of the chocolate picked	Weight eaten	Remaining weight	Result
1	30	7	[30, 25, 12]
2	25	6	[30, 25, 12]
3	12	3	[30, 25, 12]

The total weight of chocolates on day 4 is 18 + 12 = 30, which is the minimum possible weight after 4 days.

Function Description

Complete the function findMinimumTotalWeight in the editor below.

findMinimumTotalWeight has the following parameters:

• weights: an array of integers representing the weights of chocolates, indexed 0 to n - 1
• d: an integer representing the number of days

Returns

• The minimum total weight of chocolates after d days.

Constraints

• 1 ≤ n ≤ 10^5
• 1 ≤ weights[i] ≤ 10^9
• 1 ≤ d ≤ 2 * 10^9

Input Format for Custom Testing

Sample Case 0

Sample Input for Custom Testing

STDIN       Function
-----       --------
3           weights[] size n = 3
30 25 12    weights[] = [30, 25, 12]
4           d = 4

Sample Output

18

Explanation

On day 1, eat the chocolate with weight 2, and the remaining weight becomes 1. The array becomes [12] with the sum of weights 3.

On day 2, eat the chocolate with weight 2, and the remaining weight becomes 1. The array becomes [12] with the sum of weights 3.

Sample Case 1

Sample Input for Custom Testing

STDIN       Function
-----       --------
3           weights[] size n = 3
30 25 12    weights[] = [30, 25, 12]
4           d = 4

Sample Output

18

Explanation

On day 1, pick one of the chocolates.

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Maximum Order Volume from Calls

During the day, a supermarket will receive calls from customers who want to place orders. The supermarket manager knows in advance the number of calls that will be attempted, the start time, duration, and order volume for each call.

• Only one call can be in progress at any one time.
• If a call is not answered, the caller will not call back.
• The manager must choose which calls to service in order to maximize order volume.

Example

Given the following data:

Input

start = [10, 5, 15, 18, 30]
duration = [30, 12, 20, 35, 35]
volume = [50, 51, 20, 25, 10]

This can be represented as a table:

Caller	Start Time	Duration	Order Volume
1	10	30	50
2	5	12	51
3	15	20	20
4	18	35	25
5	30	35	10

Analysis

• The first call starts at 10 and lasts until 40.
• The second call starts at 5 and lasts until 17.
• The third call starts at 15 and lasts until 35.
• The fourth call starts at 18 and lasts until 53.
• The fifth call starts at 30 and lasts until 65.

The first call completely overlaps the second and third calls and partially overlaps the fourth and fifth calls.

By choosing calls that do not overlap, the optimal choice is the 2nd and 4th calls, leading to the maximum order volume of 51 + 25 = 76.

Function Description

Complete the function phoneCalls in the editor below.

Function Signature

def phoneCalls(start: List[int], duration: List[int], volume: List[int]) -> int:

Parameters

• start[n]: The start times of each call.
• duration[n]: The durations of each call.
• volume[n]: The order volumes of each call.

Returns

• int: The maximum possible order volume that can be received.

Constraints

• 1 ≤ n ≤ 10^5
• 1 ≤ start[i] ≤ 10^9

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