LeetCode #1744 — MEDIUM

Can You Eat Your Favorite Candy on Your Favorite Day?

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a (0-indexed) array of positive integers candiesCount where candiesCount[i] represents the number of candies of the i^th type you have. You are also given a 2D array queries where queries[i] = [favoriteType_i, favoriteDay_i, dailyCap_i].

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 favoriteType_i on day favoriteDay_i without eating more than dailyCap_i 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 <= favoriteType_i < candiesCount.length
0 <= favoriteDay_i <= 10⁹
1 <= dailyCap_i <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[7,4,5,3,8]
[[0,2,2],[4,2,4],[2,13,1000000000]]

Example 2

[5,2,6,4,1]
[[3,1,2],[4,10,3],[3,10,100],[4,100,30],[1,3,1]]

Step 02

Core Insight

What unlocks the optimal approach

The query is true if and only if your favorite day is in between the earliest and latest possible days to eat your favorite candy.
To get the earliest day, you need to eat dailyCap candies every day. To get the latest day, you need to eat 1 candy every day.
The latest possible day is the total number of candies with a smaller type plus the number of your favorite candy minus 1.
The earliest possible day that you can eat your favorite candy is the total number of candies with a smaller type divided by dailyCap.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1744: Can You Eat Your Favorite Candy on Your Favorite Day?
class Solution {
    public boolean[] canEat(int[] candiesCount, int[][] queries) {
        int n = candiesCount.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + candiesCount[i];
        }
        int m = queries.length;
        boolean[] ans = new boolean[m];
        for (int i = 0; i < m; ++i) {
            int t = queries[i][0], day = queries[i][1], mx = queries[i][2];
            long least = day, most = (long) (day + 1) * mx;
            ans[i] = least < s[t + 1] && most > s[t];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1744: Can You Eat Your Favorite Candy on Your Favorite Day?
func canEat(candiesCount []int, queries [][]int) (ans []bool) {
	n := len(candiesCount)
	s := make([]int, n+1)
	for i, v := range candiesCount {
		s[i+1] = s[i] + v
	}
	for _, q := range queries {
		t, day, mx := q[0], q[1], q[2]
		least, most := day, (day+1)*mx
		ans = append(ans, least < s[t+1] && most > s[t])
	}
	return
}

# Accepted solution for LeetCode #1744: Can You Eat Your Favorite Candy on Your Favorite Day?
class Solution:
    def canEat(self, candiesCount: List[int], queries: List[List[int]]) -> List[bool]:
        s = list(accumulate(candiesCount, initial=0))
        ans = []
        for t, day, mx in queries:
            least, most = day, (day + 1) * mx
            ans.append(least < s[t + 1] and most > s[t])
        return ans

// Accepted solution for LeetCode #1744: Can You Eat Your Favorite Candy on Your Favorite Day?
struct Solution;

impl Solution {
    fn can_eat(candies_count: Vec<i32>, queries: Vec<Vec<i32>>) -> Vec<bool> {
        let mut prefix: Vec<i64> = vec![0];
        let mut prev = 0;
        for count in candies_count {
            prev += count as i64;
            prefix.push(prev);
        }
        let mut res = vec![];
        for q in queries {
            let t = q[0] as usize;
            let d = q[1] as i64;
            let c = q[2] as i64;
            res.push(prefix[t] / c <= d && d < prefix[t + 1]);
        }

        res
    }
}

#[test]
fn test() {
    let candies_count = vec![7, 4, 5, 3, 8];
    let queries = vec_vec_i32![[0, 2, 2], [4, 2, 4], [2, 13, 1000000000]];
    let res = vec![true, false, true];
    assert_eq!(Solution::can_eat(candies_count, queries), res);
    let candies_count = vec![5, 2, 6, 4, 1];
    let queries = vec_vec_i32![[3, 1, 2], [4, 10, 3], [3, 10, 100], [4, 100, 30], [1, 3, 1]];
    let res = vec![false, true, true, false, false];
    assert_eq!(Solution::can_eat(candies_count, queries), res);
    let candies_count = vec![7, 11, 5, 3, 8];
    let queries = vec_vec_i32![[2, 2, 6], [4, 2, 4], [2, 13, 1000000000]];
    let res = vec![false, false, true];
    assert_eq!(Solution::can_eat(candies_count, queries), res);
}

// Accepted solution for LeetCode #1744: Can You Eat Your Favorite Candy on Your Favorite Day?
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1744: Can You Eat Your Favorite Candy on Your Favorite Day?
// class Solution {
//     public boolean[] canEat(int[] candiesCount, int[][] queries) {
//         int n = candiesCount.length;
//         long[] s = new long[n + 1];
//         for (int i = 0; i < n; ++i) {
//             s[i + 1] = s[i] + candiesCount[i];
//         }
//         int m = queries.length;
//         boolean[] ans = new boolean[m];
//         for (int i = 0; i < m; ++i) {
//             int t = queries[i][0], day = queries[i][1], mx = queries[i][2];
//             long least = day, most = (long) (day + 1) * mx;
//             ans[i] = least < s[t + 1] && most > s[t];
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.