Introduction

In this blog post, we will explore how to solve the LeetCode problem "1744" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a (0-indexed) array of positive integers candiesCount where candiesCount[i] represents the number of candies of the ith type you have. You are also given a 2D array queries where queries[i] = [favoriteTypei, favoriteDayi, dailyCapi]. You play a game with the following rules: You start eating candies on day 0. You cannot eat any candy of type i unless you have eaten all candies of type i - 1. You must eat at least one candy per day until you have eaten all the candies. Construct a boolean array answer such that answer.length == queries.length and answer[i] is true if you can eat a candy of type favoriteTypei on day favoriteDayi without eating more than dailyCapi candies on any day, and false otherwise. Note that you can eat different types of candy on the same day, provided that you follow rule 2. Return the constructed array answer. Example 1: Input: candiesCount = [7,4,5,3,8], queries = [[0,2,2],[4,2,4],[2,13,1000000000]] Output: [true,false,true] Explanation: 1- If you eat 2 candies (type 0) on day 0 and 2 candies (type 0) on day 1, you will eat a candy of type 0 on day 2. 2- You can eat at most 4 candies each day. If you eat 4 candies every day, you will eat 4 candies (type 0) on day 0 and 4 candies (type 0 and type 1) on day 1. On day 2, you can only eat 4 candies (type 1 and type 2), so you cannot eat a candy of type 4 on day 2. 3- If you eat 1 candy each day, you will eat a candy of type 2 on day 13. Example 2: Input: candiesCount = [5,2,6,4,1], queries = [[3,1,2],[4,10,3],[3,10,100],[4,100,30],[1,3,1]] Output: [false,true,true,false,false] Constraints: 1 <= candiesCount.length <= 105 1 <= candiesCount[i] <= 105 1 <= queries.length <= 105 queries[i].length == 3 0 <= favoriteTypei < candiesCount.length 0 <= favoriteDayi <= 109 1 <= dailyCapi <= 109

Explanation

Here's a breakdown of the solution:

• Prefix Sums for Efficient Calculation: Calculate prefix sums of candiesCount to determine the total number of candies available up to a certain type. This allows for constant-time calculation of the minimum and maximum possible days to reach a certain candy type.

• Check Feasibility: For each query, determine the earliest and latest days a candy of the desired type could be eaten. The earliest day is determined by the number of candies before the target type. The latest day depends on the daily capacity and number of candies of the target type.

• Boolean Array Creation: Store True if the favoriteDayi falls within the feasible range and False otherwise.

• Runtime Complexity: O(n + q), where n is the length of candiesCount, and q is the number of queries. This is because we iterate through candiesCount once to compute the prefix sum and iterate through queries once.

• Storage Complexity: O(n + q), due to the prefix sum array and the result boolean array.

Code

    def canEat(candiesCount, queries):
    n = len(candiesCount)
    prefix_sum = [0] * (n + 1)
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + candiesCount[i]

    result = []
    for favoriteType, favoriteDay, dailyCap in queries:
        min_day = prefix_sum[favoriteType] // dailyCap
        max_day = (prefix_sum[favoriteType + 1] - 1) 

        result.append(min_day <= favoriteDay <= max_day)

    return result