LeetCode #3679 — MEDIUM

Minimum Discards to Balance Inventory

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

The Problem

Problem Statement

You are given two integers w and m, and an integer array arrivals, where arrivals[i] is the type of item arriving on day i (days are 1-indexed).

Items are managed according to the following rules:

Each arrival may be kept or discarded; an item may only be discarded on its arrival day.
For each day i, consider the window of days [max(1, i - w + 1), i] (the w most recent days up to day i):
- For any such window, each item type may appear at most m times among kept arrivals whose arrival day lies in that window.
- If keeping the arrival on day i would cause its type to appear more than m times in the window, that arrival must be discarded.

Return the minimum number of arrivals to be discarded so that every w-day window contains at most m occurrences of each type.

Example 1:

Input: arrivals = [1,2,1,3,1], w = 4, m = 2

Output: 0

Explanation:

On day 1, Item 1 arrives; the window contains no more than m occurrences of this type, so we keep it.
On day 2, Item 2 arrives; the window of days 1 - 2 is fine.
On day 3, Item 1 arrives, window [1, 2, 1] has item 1 twice, within limit.
On day 4, Item 3 arrives, window [1, 2, 1, 3] has item 1 twice, allowed.
On day 5, Item 1 arrives, window [2, 1, 3, 1] has item 1 twice, still valid.

There are no discarded items, so return 0.

Example 2:

Input: arrivals = [1,2,3,3,3,4], w = 3, m = 2

Output: 1

Explanation:

On day 1, Item 1 arrives. We keep it.
On day 2, Item 2 arrives, window [1, 2] is fine.
On day 3, Item 3 arrives, window [1, 2, 3] has item 3 once.
On day 4, Item 3 arrives, window [2, 3, 3] has item 3 twice, allowed.
On day 5, Item 3 arrives, window [3, 3, 3] has item 3 three times, exceeds limit, so the arrival must be discarded.
On day 6, Item 4 arrives, window [3, 4] is fine.

Item 3 on day 5 is discarded, and this is the minimum number of arrivals to discard, so return 1.

Constraints:

1 <= arrivals.length <= 10⁵
1 <= arrivals[i] <= 10⁵
1 <= w <= arrivals.length
1 <= m <= w

Step 01

Brute Force Baseline

Problem summary: You are given two integers w and m, and an integer array arrivals, where arrivals[i] is the type of item arriving on day i (days are 1-indexed). Items are managed according to the following rules: Each arrival may be kept or discarded; an item may only be discarded on its arrival day. For each day i, consider the window of days [max(1, i - w + 1), i] (the w most recent days up to day i): For any such window, each item type may appear at most m times among kept arrivals whose arrival day lies in that window. If keeping the arrival on day i would cause its type to appear more than m times in the window, that arrival must be discarded. Return the minimum number of arrivals to be discarded so that every w-day window contains at most m occurrences of each type.

Baseline thinking

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

Pattern signal: Array · Hash Map · Sliding Window

Example 1

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

Example 2

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

Step 02

Core Insight

What unlocks the optimal approach

Use a sliding window of up to <code>w</code> days with two pointers <code>left</code> and <code>right</code> to represent the current interval.
Maintain a hash map <code>cnt</code> from item type to its current count in the window. When you advance <code>right</code> to day <code>i</code>, do <code>cnt[arrivals[i]]++</code>.
If the window size exceeds <code>w</code> (i.e. <code>right - left + 1 > w</code>), shrink it by doing <code>cnt[arrivals[left]]--</code> and then <code>left++</code>.
After each increment, check if <code>cnt[arrivals[right]] > m</code>. If so, we must discard the current arrival.

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 #3679:  Minimum Discards to Balance Inventory
class Solution {
    public int minArrivalsToDiscard(int[] arrivals, int w, int m) {
        Map<Integer, Integer> cnt = new HashMap<>();
        int n = arrivals.length;
        int[] marked = new int[n];
        int ans = 0;
        for (int i = 0; i < n; i++) {
            int x = arrivals[i];
            if (i >= w) {
                int prev = arrivals[i - w];
                cnt.merge(prev, -marked[i - w], Integer::sum);
            }
            if (cnt.getOrDefault(x, 0) >= m) {
                ans++;
            } else {
                marked[i] = 1;
                cnt.merge(x, 1, Integer::sum);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3679:  Minimum Discards to Balance Inventory
func minArrivalsToDiscard(arrivals []int, w int, m int) (ans int) {
	cnt := make(map[int]int)
	n := len(arrivals)
	marked := make([]int, n)
	for i, x := range arrivals {
		if i >= w {
			cnt[arrivals[i-w]] -= marked[i-w]
		}
		if cnt[x] >= m {
			ans++
		} else {
			marked[i] = 1
			cnt[x]++
		}
	}
	return
}

# Accepted solution for LeetCode #3679:  Minimum Discards to Balance Inventory
class Solution:
    def minArrivalsToDiscard(self, arrivals: List[int], w: int, m: int) -> int:
        cnt = Counter()
        n = len(arrivals)
        marked = [0] * n
        ans = 0
        for i, x in enumerate(arrivals):
            if i >= w:
                cnt[arrivals[i - w]] -= marked[i - w]
            if cnt[x] >= m:
                ans += 1
            else:
                marked[i] = 1
                cnt[x] += 1
        return ans

// Accepted solution for LeetCode #3679:  Minimum Discards to Balance Inventory
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3679:  Minimum Discards to Balance Inventory
// class Solution {
//     public int minArrivalsToDiscard(int[] arrivals, int w, int m) {
//         Map<Integer, Integer> cnt = new HashMap<>();
//         int n = arrivals.length;
//         int[] marked = new int[n];
//         int ans = 0;
//         for (int i = 0; i < n; i++) {
//             int x = arrivals[i];
//             if (i >= w) {
//                 int prev = arrivals[i - w];
//                 cnt.merge(prev, -marked[i - w], Integer::sum);
//             }
//             if (cnt.getOrDefault(x, 0) >= m) {
//                 ans++;
//             } else {
//                 marked[i] = 1;
//                 cnt.merge(x, 1, Integer::sum);
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3679:  Minimum Discards to Balance Inventory
function minArrivalsToDiscard(arrivals: number[], w: number, m: number): number {
    const cnt = new Map<number, number>();
    const n = arrivals.length;
    const marked = Array<number>(n).fill(0);
    let ans = 0;

    for (let i = 0; i < n; i++) {
        const x = arrivals[i];
        if (i >= w) {
            cnt.set(arrivals[i - w], (cnt.get(arrivals[i - w]) || 0) - marked[i - w]);
        }
        if ((cnt.get(x) || 0) >= m) {
            ans++;
        } else {
            marked[i] = 1;
            cnt.set(x, (cnt.get(x) || 0) + 1);
        }
    }
    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(n)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.