LeetCode #1705 — MEDIUM

Maximum Number of Eaten Apples

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

The Problem

Problem Statement

There is a special kind of apple tree that grows apples every day for n days. On the i^th day, the tree grows apples[i] apples that will rot after days[i] days, that is on day i + days[i] the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by apples[i] == 0 and days[i] == 0.

You decided to eat at most one apple a day (to keep the doctors away). Note that you can keep eating after the first n days.

Given two integer arrays days and apples of length n, return the maximum number of apples you can eat.

Example 1:

Input: apples = [1,2,3,5,2], days = [3,2,1,4,2]
Output: 7
Explanation: You can eat 7 apples:
- On the first day, you eat an apple that grew on the first day.
- On the second day, you eat an apple that grew on the second day.
- On the third day, you eat an apple that grew on the second day. After this day, the apples that grew on the third day rot.
- On the fourth to the seventh days, you eat apples that grew on the fourth day.

Example 2:

Input: apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
Output: 5
Explanation: You can eat 5 apples:
- On the first to the third day you eat apples that grew on the first day.
- Do nothing on the fouth and fifth days.
- On the sixth and seventh days you eat apples that grew on the sixth day.

Constraints:

n == apples.length == days.length
1 <= n <= 2 * 10⁴
0 <= apples[i], days[i] <= 2 * 10⁴
days[i] = 0 if and only if apples[i] = 0.

Step 01

Brute Force Baseline

Problem summary: There is a special kind of apple tree that grows apples every day for n days. On the ith day, the tree grows apples[i] apples that will rot after days[i] days, that is on day i + days[i] the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by apples[i] == 0 and days[i] == 0. You decided to eat at most one apple a day (to keep the doctors away). Note that you can keep eating after the first n days. Given two integer arrays days and apples of length n, return the maximum number of apples you can eat.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

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

Example 2

[3,0,0,0,0,2]
[3,0,0,0,0,2]

Step 02

Core Insight

What unlocks the optimal approach

It's optimal to finish the apples that will rot first before those that will rot last
You need a structure to keep the apples sorted by their finish time

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 #1705: Maximum Number of Eaten Apples
class Solution {
    public int eatenApples(int[] apples, int[] days) {
        PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        int n = days.length;
        int ans = 0, i = 0;
        while (i < n || !q.isEmpty()) {
            if (i < n && apples[i] > 0) {
                q.offer(new int[] {i + days[i] - 1, apples[i]});
            }
            while (!q.isEmpty() && q.peek()[0] < i) {
                q.poll();
            }
            if (!q.isEmpty()) {
                var p = q.poll();
                ++ans;
                if (--p[1] > 0 && p[0] > i) {
                    q.offer(p);
                }
            }
            ++i;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1705: Maximum Number of Eaten Apples
func eatenApples(apples []int, days []int) int {
	var h hp
	ans, n := 0, len(apples)
	for i := 0; i < n || len(h) > 0; i++ {
		if i < n && apples[i] > 0 {
			heap.Push(&h, pair{i + days[i] - 1, apples[i]})
		}
		for len(h) > 0 && h[0].first < i {
			heap.Pop(&h)
		}
		if len(h) > 0 {
			h[0].second--
			if h[0].first == i || h[0].second == 0 {
				heap.Pop(&h)
			}
			ans++
		}
	}
	return ans
}

type pair struct {
	first  int
	second int
}

type hp []pair

func (a hp) Len() int           { return len(a) }
func (a hp) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a hp) Less(i, j int) bool { return a[i].first < a[j].first }
func (a *hp) Push(x any)        { *a = append(*a, x.(pair)) }
func (a *hp) Pop() any          { l := len(*a); t := (*a)[l-1]; *a = (*a)[:l-1]; return t }

# Accepted solution for LeetCode #1705: Maximum Number of Eaten Apples
class Solution:
    def eatenApples(self, apples: List[int], days: List[int]) -> int:
        n = len(days)
        i = ans = 0
        q = []
        while i < n or q:
            if i < n and apples[i]:
                heappush(q, (i + days[i] - 1, apples[i]))
            while q and q[0][0] < i:
                heappop(q)
            if q:
                t, v = heappop(q)
                v -= 1
                ans += 1
                if v and t > i:
                    heappush(q, (t, v))
            i += 1
        return ans

// Accepted solution for LeetCode #1705: Maximum Number of Eaten Apples
struct Solution;
use std::cmp::Reverse;
use std::collections::BinaryHeap;

impl Solution {
    fn eaten_apples(apples: Vec<i32>, days: Vec<i32>) -> i32 {
        let n = apples.len();
        let mut queue: BinaryHeap<Reverse<(usize, i32)>> = BinaryHeap::new();
        let mut i = 0;
        let mut res = 0;
        loop {
            if i < n {
                queue.push(Reverse((i + days[i] as usize, apples[i])));
            }
            if i >= n && queue.is_empty() {
                break;
            }
            while let Some(Reverse((j, left))) = queue.peek() {
                if i >= *j || *left == 0 {
                    queue.pop();
                } else {
                    break;
                }
            }
            if let Some(mut top) = queue.peek_mut() {
                (top.0).1 -= 1;
                res += 1;
            }
            i += 1;
        }
        res
    }
}

#[test]
fn test() {
    let apples = vec![1, 2, 3, 5, 2];
    let days = vec![3, 2, 1, 4, 2];
    let res = 7;
    assert_eq!(Solution::eaten_apples(apples, days), res);
    let apples = vec![3, 0, 0, 0, 0, 2];
    let days = vec![3, 0, 0, 0, 0, 2];
    let res = 5;
    assert_eq!(Solution::eaten_apples(apples, days), res);
}

// Accepted solution for LeetCode #1705: Maximum Number of Eaten Apples
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1705: Maximum Number of Eaten Apples
// class Solution {
//     public int eatenApples(int[] apples, int[] days) {
//         PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
//         int n = days.length;
//         int ans = 0, i = 0;
//         while (i < n || !q.isEmpty()) {
//             if (i < n && apples[i] > 0) {
//                 q.offer(new int[] {i + days[i] - 1, apples[i]});
//             }
//             while (!q.isEmpty() && q.peek()[0] < i) {
//                 q.poll();
//             }
//             if (!q.isEmpty()) {
//                 var p = q.poll();
//                 ++ans;
//                 if (--p[1] > 0 && p[0] > i) {
//                     q.offer(p);
//                 }
//             }
//             ++i;
//         }
//         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 × log n + M)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.