Problem Description

In this problem, we are required to simulate a game of eating candies with some specific rules:

1. We are given an array candiesCount where each element candiesCount[i] denotes the number of candies of type i.
2. We also have a list of queries. Each query is given as an array with 3 elements: [favoriteType, favoriteDay, dailyCap].
3. The game starts on day 0 and we must eat at least one candy every day.
4. We must finish eating all candies of type i before we can start on type i+1.
5. Our goal is to determine for each query if it is possible to eat at least one candy of our favoriteType on the favoriteDay without exceeding a maximum number of candies defined by dailyCap.

The essence of the problem is to check if our eating strategy, adhering to the rules, allows us to satisfy all the conditions described in a query.

Intuition

The intuition behind the solution leverages the understanding that to satisfy the queries, we must figure out the minimum and maximum number of candies we could have eaten by favoriteDay. These two values create a range, and if our favoriteType candy falls within that range, the answer for the query is true. Otherwise, it's false.

Here's how we arrive at the solution:

1. First, we use prefix sums to calculate the total number of candies eaten up to each type. Prefix sums help us quickly determine the total count without repeatedly summing individual counts.
2. For each query, we calculate the minimum number of candies possibly eaten by favoriteDay by assuming we eat only one candy per day, which is the same as the favoriteDay itself because we start at day 0.
3. We also calculate the maximum number of candies by multiplying the favoriteDay plus one (since we start at day 0) by the dailyCap.
4. The favoriteType candy can be eaten if it falls within the range created by our minimum and maximum possible candies. This means the following two conditions must hold true:
   • The number of candies eaten up to favoriteType - 1 (thus s[favoriteType]) must be less than the most we could eat by favoriteDay (which we've calculated).
   • And, the number of candies we could have started eating by favoriteDay (least + 1) must be less than or equal to the total candy count up to including favoriteType (s[favoriteType + 1]).

The given solution neatly calculates this for every query, recording the answers in an array ans to return the results.

Learn more about Prefix Sum patterns.

Solution Approach

The solution is implemented in Python using the following concepts:

1. Prefix Sums: We use the accumulate function from Python's itertools module to generate a list s that contains the cumulative sum of the candies. The initial=0 parameter ensures that there is a 0 prefixed to the accumulated list, which facilitates calculating the number of candies eaten before a certain type.

2. Array Manipulation: The primary operation in the solution is element-wise manipulation and comparison within arrays or lists. These include:

   • Accessing the sums of candies before the current type (s[t]).
   • Calculating the sums of candies including the current type (s[t + 1]).

3. Iterating Over Queries: Each query is represented as a list [t, day, mx], and we loop through the queries list while calculating the corresponding boolean value for each query to determine if the eating plan for a given favoriteType is feasible on the favoriteDay.

4. Calculating Minimum and Maximum Candy Range: For every query, we calculate the minimum (least) and maximum (most) number of candies that could have been eaten by the favoriteDay.

   • least is the day number itself, which assumes that we eat one candy per day.
   • most is the number (day + 1) * mx, reflecting the maximum consumption rate according to the dailyCap.

5. Range Checking: Using the values in s, we check if least is less than s[t+1], which signifies that the least number of candies we could eat by favoriteDay is enough to allow us to start eating the favoriteType. We also check that most is greater than s[t], which ensures we haven't already surpassed the amount of the favoriteType of candy.

6. Appending Results: If both conditions hold true for a query, we append True to our ans list; otherwise, we append False. This list is what is eventually returned.

The implementation can be summarized in the following steps:

• Calculate the prefix sums of candiesCount and store it in s.
• For each query in queries list:
  • Extract favoriteType (t), favoriteDay (day), and dailyCap (mx) from the query.
  • Calculate the least and most number of candies that could have been eaten by the favoriteDay.
  • Append True or False to the ans list based on if the ranges overlap with the amount available for the favoriteType.
• Return the ans list once all queries have been processed.

Example Walkthrough

Let's go through a small example to illustrate the solution approach provided.

Suppose we have an input array, candiesCount, and a list of queries given as follows:

candiesCount = [7, 4, 5, 3]
queries = [[0, 2, 2], [1, 6, 1], [2, 5, 10]]

In this scenario:

• We have 7 candies of type 0, 4 of type 1, 5 of type 2, and 3 of type 3.
• The queries detail scenarios where we want to know if we can eat our favorite type of candy on the favorite day without exceeding a set daily cap.

Calculate Prefix Sums

Firstly, we calculate prefix sums of candiesCount:

s = [0, 7, 11, 16, 19]  # Calculated using prefix sums; we added a 0 at the start.

Process Each Query

Now, let's process each query using our example:

Query 1: [0, 2, 2]

• Favorite type: 0
• Favorite day: 2
• Daily cap: 2

We calculate the least number of candies possibly eaten by day 2, which is 2, and the most, which is (2 + 1) * 2 = 6.

Since s[0] = 0 and s[1] = 7, the number of candies of type 0 we could have started eating by day 3 is less than 7. Thus, 0 < 7 <= 6 does not hold true, so our answer is False.

Query 2: [1, 6, 1]

• Favorite type: 1
• Favorite day: 6
• Daily cap: 1

Calculating least as 6 and most as (6 + 1) * 1 = 7.

The number of candies eaten before type 1 (s[1] = 7) should be less than most = 7, and the number we can start eating by day 6 (least + 1 = 7) should be less than or equal to the candies of type 1 available (s[2] = 11), which means 7 <= 11. Now, 6 < 11 and 7 <= 11 both hold true, hence the response is True.

Query 3: [2, 5, 10]

• Favorite type: 2
• Favorite day: 5
• Daily cap: 10

least is 5, and most now is (5 + 1) * 10 = 60.

Considering s[2] = 11 and s[3] = 16, we have not eaten type 2 candies by day 5, since 11 <= 60. We could also start eating them the same day since 5 + 1 = 6 is less than 16. Thus, the conditions 11 <= 60 and 6 <= 16 are satisfied, resulting in True.

Construct Answer Array

Finally, we have processed all queries and our answer array ans is [False, True, True], indicating the feasibility of each query.

Complete Process

To apply this approach in an actual Python solution, follow the summarized steps outlined in the content provided. Calculate the prefix sums, iterate over every query to determine the range in which we can eat candies, check the range conditions, and then append the result to the answer list which is then returned.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code is O(n + q), where n is the length of the candiesCount array, and q is the number of queries. The accumulate() function is called once on the candiesCount array, and this operation has a time complexity of O(n) for calculating the prefix sums. Then the algorithm processes each query in constant time, so the time complexity for processing all queries is O(q). Therefore, the combined time complexity is the sum of the two, which results in O(n + q).

The space complexity of the provided code is O(n), as the s array stores the prefix sums of the candiesCount array, which is the only additional significant space used by the algorithm.

Please note that the space taken by the output array ans is not included in this analysis, as it's usually considered the space required to store the output of the function, and is typically not counted towards the additional space complexity of an algorithm.

Learn more about how to find time and space complexity quickly using problem constraints.