1. Minimum Total Weight

You have n chocolates with weights given in an array weights[n], where weights[i] represents the weight of the iᵗʰ chocolate. Each day, you can pick one chocolate, consume half of its weight (calculated as floor(weights[i] / 2)), and keep the remaining portion. Calculate the minimum possible total weight of the chocolates after d days. Note that you can eat from the same chocolate multiple times.

Example:

weights = [30, 20, 25]
d = 4

Day	Weight of the chocolate picked	Weight eaten	Remaining weight	Result
1	20	10	10	[30, 10, 25]
2	25	12	13	[30, 10, 13]
3	30	15	15	[15, 10, 13]
4	15	7	8	[8, 10, 13]

The total weight of chocolates on day 4 is 8 + 10 + 13 = 31, which is the minimum possible weight after 4 days.

Function Description

Complete the function findMinWeight in the editor below.

Function Signature:

int findMinWeight(int weights[], int d)

Parameters:

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

Returns:

• int: the minimum total weight of chocolates after d days.

Constraints

• 1 ≤ n ≤ 10⁵
• 1 ≤ weights[i] ≤ 10⁴ (where 0 ≤ i < n)
• 1 ≤ d ≤ 2 * 10⁶

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2. Paint the Ceiling

You want to build yourself a house. The building company you hired can only build houses with sides from their specific set s. That means you can build a square house or a rectangular one, but only if its length and width belong to the set s.

This month, they have a special promotion: they will paint the ceiling of a new house for free, but only if its area is not more than a. You want them to do it for free, but you also want to be sure that the house will be comfortable and not too small. How many possible configurations can you create to have the ceiling painted for free given the side lengths offered?

Generating the Side Lengths

There is a method to determine what lengths of sides are available. To determine n lengths of wall segments, they start with a seed value s₀, and use variables k, b, m to generate all other side lengths s[i] using the formula:

s[i] = ((k * s[i - 1] + b) mod m) + s[i - 1], for 1 ≤ i < n

For example, given s₀ = 2, and n = 3, using k = 3, b = 3, m = 2, we calculate:

s	i	Calculation	Result
[2]	1	-	2
[2, 4]	2	((3 * 2 + 3) % 2) + 2	4
[2, 4, 6]	3	((3 * 4 + 3) % 2) + 4	6

Now that we have the set of available lengths, we need to determine how many valid house configurations exist.

Finding Valid House Configurations

If we assume a = 15, we can brute force possible house configurations:

s₁	s₂	s₁ * s₂	s₁ * s₂ ≤ a
2	2	4	True
2	4	8	True
2	6	12	True
4	2	8	True
4	4	16	False
4	6	24	False
6	2	12	True
6	4	24	False
6	6	36	False

There are 5 valid configurations that result in a free paint job.

Function Description

Complete the function variantsCount in the editor below. The function must return an integer that denotes the number of valid variants that allow you to use the promotion.

Function Signature:

int variantsCount(int n, int s0, int k, int b, int m, long a)

Parameters:

• n: an integer, the number of wall lengths offered
• s0: an integer, the length of the shortest wall
• k, b, m: three arbitrary integers
• a: a long integer, the largest area that will be painted for free

Constraints

• 1 ≤ n ≤ 2 * 10⁷
• 1 ≤ s[i] ≤ 10⁹ for 0 ≤ i < n
• 1 ≤ k, b, m ≤ 10⁹
• 1 ≤ a ≤ 10¹⁸

Sample Case

Sample Input:

3
1
1
1
2
4

Sample Output:

6

Explanation:

With n = 3, s0 = 1, k = 1, b = 1, m = 2, and a = 4, we generate:

i	s[i] Calculation	s[i]
0	Given: s₀	1
1	((1 * 1 + 1) % 2) + 1	2
2	((1 * 2 + 1) % 2) + 2	4

Possible configurations:

s₁	s₂	Area (s₁ * s₂)	≤ a
1	1	1	✅
1	2	2	✅
1	4	4	✅
2	2	4	✅
2	4	8	❌
4	4	16	❌

A total of 6 valid configurations exist where the area is ≤ 4.

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3. Maximum Order Volume

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, and 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. Determine the maximum order volume.

Example

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

The above data 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

• The first call will start at time 10 and last until 40.
• The second call will start at time 5 and last until 17.
• The third call will start at time 15 and last until 35.
• The fourth call will start at time 18 and last until 53.
• The fifth call will start at time 30 and last until 65.

The first call completely overlaps the second and third calls and partially overlaps the fourth and fifth calls. Choosing calls that do not overlap and answering the 2nd and 4th calls leads to the maximum total order volume of:

51 + 25 = 76

Function Description

Complete the function phoneCalls in the editor below.

Function Signature:

int phoneCalls(int start[], int duration[], int volume[], int n)

Parameters:

• start[n]: an array of integers representing the start times of each call
• duration[n]: an array of integers representing the durations of each call
• volume[n]: an array of integers representing the volumes of each order
• n: an integer, the number of calls

Constraints

• 1 ≤ n ≤ 10⁵
• 1 ≤ start[i] ≤ 10⁹
• 1 ≤ duration[i] ≤ 10⁹
• 1 ≤ volume[i] ≤ 10⁹

Sample Case 1

Sample Input:

3
start = [1, 2, 4]
duration = [2, 2, 1]
volume = [1, 2, 3]

Sample Output:

4

Explanation:

The calls happen in the following intervals:

Call	Interval
1	[1, 3]
2	[2, 4]
3	[4, 5]

• The first and third calls together make up the order volume 4, and their intervals do not intersect.
• The first and second calls intersect, as do the second and third calls.
• Only one call from either of these pairs can be serviced.
• The most efficient calls to answer are the first and third, with a total volume of 4.

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Sample Case 2

Sample Input:

3
start = [1, 10, 100]
duration = [1, 10, 100]
volume = [1, 10, 100]

Sample Output:

111

Explanation:

The calls happen in the following intervals:

Call	Interval
1	[1, 2]
2	[10, 20]
3	[100, 200]

• All three calls can be attended, so the total volume is:

1 + 10 + 100 = 111

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