1744. Can You Eat Your Favorite Candy on Your Favorite Day

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.


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.

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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.
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1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
1import java.util.StringJoiner;
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);

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:

1candiesCount = [7, 4, 5, 3]
2queries = [[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:

1s = [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.

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Python Solution

1from itertools import accumulate
3class Solution:
4    def canEat(self, candies_count: List[int], queries: List[List[int]]) -> List[bool]:
5        # Use accumulate function with initial=0 to create a prefix sum array
6        # where s[i] represents the total number of candies up to (but not including) index i
7        prefix_sum = list(accumulate(candies_count, initial=0))
9        # Initialize an empty list to store the answer to each query
10        answer_list = []
12        # Iterate through each query in the queries list
13        for candy_type, day, max_candies_per_day in queries:
14            # Calculate the least number of candies the user could eat
15            least_candies_eaten = day
16            # Calculate the most number of candies the user could eat
17            most_candies_eaten = (day + 1) * max_candies_per_day
19            # Determine if the user can eat at least one candy of type 'candy_type'
20            # by checking if they would still have candies left on the 'day'
21            # and also making sure they won't run out of candies before the day
22            can_eat = least_candies_eaten < prefix_sum[candy_type + 1] and most_candies_eaten > prefix_sum[candy_type]
24            # Add the result to the answer list
25            answer_list.append(can_eat)
27        # Return the list of answers for each query
28        return answer_list

Java Solution

1class Solution {
2    public boolean[] canEat(int[] candiesCount, int[][] queries) {
3        int typesOfCandies = candiesCount.length;
4        long[] cumulativeCandies = new long[typesOfCandies + 1];
6        // Calculate the cumulative sum of candies to determine the total
7        // number of candies up to a certain type
8        for (int i = 0; i < typesOfCandies; ++i) {
9            cumulativeCandies[i + 1] = cumulativeCandies[i] + candiesCount[i];
10        }
12        int numberOfQueries = queries.length;
13        boolean[] canEatOnDay = new boolean[numberOfQueries];
15        // Iterate over each query to determine if you can eat the candies
16        for (int i = 0; i < numberOfQueries; ++i) {
17            int type = queries[i][0]; // Type of candy
18            int day = queries[i][1]; // Day number
19            int maxCandies = queries[i][2]; // Maximum number of candies you can eat
21            // The earliest amount of candies you could eat by that day
22            // is equal to the day number if you eat 1 candy per day
24            // Whereas the most amount of candies you could eat is if
25            // you eat 'maxCandies' on each day including 'day'
27            long leastCandiesYouCouldEat = day; 
28            long mostCandiesYouCouldEat = (long) (day + 1) * maxCandies;
30            // Check if there's any overlap between the range of candies you could eat
31            // and the range of candies available for that type. You can eat the candy
32            // on that day if at least one candy of that type is within your eating range.
33            canEatOnDay[i] = leastCandiesYouCouldEat < cumulativeCandies[type + 1] && mostCandiesYouCouldEat > cumulativeCandies[type];
34        }
36        return canEatOnDay;
37    }

C++ Solution

1#include <vector>
3// Type alias for long long for ease of use
4using ll = long long;
6class Solution {
8    // Function to determine if you can eat all your favorite candies
9    vector<bool> canEat(vector<int>& candiesCount, vector<vector<int>>& queries) {
11        // Get the number of different types of candies
12        int numTypes = candiesCount.size();
14        // Prefix sum array to store the total candies up to index i
15        vector<ll> prefixSum(numTypes + 1);
16        for (int i = 0; i < numTypes; ++i) 
17            prefixSum[i + 1] = prefixSum[i] + candiesCount[i];
19        // Result vector to store whether each query is true or false
20        vector<bool> results;
22        // Process each query
23        for (auto& query : queries) {
24            // Type of favorite candy for this query
25            int favoriteType = query[0];
26            // Day on which you plan to eat the favorite candy
27            int day = query[1];
28            // Maximum number of candies you can eat each day
29            int maxCandiesPerDay = query[2];
31            // Least number of candies you could have eaten until 'day'
32            ll leastCandiesEaten = day;
33            // Most number of candies you could have eaten until 'day'
34            ll mostCandiesEaten = 1ll * (day + 1) * maxCandiesPerDay;
36            // Check if it is possible to eat some favorite candies on that day
37            // We can eat our favorite candy only if we have not eaten all of them before that day
38            // and we would have been able to reach to our favorite candy type by that day.
39            results.emplace_back(leastCandiesEaten < prefixSum[favoriteType + 1] && 
40                                 mostCandiesEaten > prefixSum[favoriteType]);
41        }
43        // Return result vector
44        return results;
45    }

Typescript Solution

1// Define a type alias for number to represent large counts
2type ll = number;
4// Array to store counts for each type of candy
5let candiesCount: number[];
7// Matrix to store queries, where each query is an array containing:
8// 0. Type of favorite candy
9// 1. Day to eat the favorite candy
10// 2. Maximum number of candies that can be eaten in one day
11let queries: number[][];
14 * Function to determine if you can eat all your favorite candies based on queries
15 * @param candiesCount - Array with count of each type of candy
16 * @param queries - Array of queries with details [favoriteType, day, maxCandiesPerDay]
17 * @returns Array of boolean values where each value corresponds to the possibility of a query
18 */
19function canEat(candiesCount: number[], queries: number[][]): boolean[] {
21    // Get the number of different types of candies
22    const numTypes: number = candiesCount.length;
24    // Array to store the prefix sum for total candies up to index i
25    const prefixSum: ll[] = new Array(numTypes + 1).fill(0);
26    for (let i = 0; i < numTypes; ++i) {
27        prefixSum[i + 1] = prefixSum[i] + candiesCount[i];
28    }
30    // Array to store the results of the queries
31    const results: boolean[] = [];
33    // Iterate through each query and process it
34    for (const query of queries) {
35        // Extract the query details for readability
36        const favoriteType: number = query[0];
37        const day: number = query[1];
38        const maxCandiesPerDay: number = query[2];
40        // Calculate the least number of candies that could have been eaten until the 'day'
41        const leastCandiesEaten: ll = day;
43        // Calculate the most number of candies that could have been eaten until the 'day'
44        const mostCandiesEaten: ll = (day + 1) * maxCandiesPerDay;
46        // Determine if it's possible to eat some of the favorite candies on that day
47        const canEatFavorite: boolean = leastCandiesEaten < prefixSum[favoriteType + 1]
48                                        && mostCandiesEaten > prefixSum[favoriteType];
50        // Add the result to the results array
51        results.push(canEatFavorite);
52    }
54    // Return the array of results from the queries
55    return results;
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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.

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