LeetCode-in-Java

3679. Minimum Discards to Balance Inventory

Medium

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

Solution

public class Solution {
    public int minArrivalsToDiscard(int[] arrivals, int w, int m) {
        int n = arrivals.length;
        int dis = 0;
        boolean[] removed = new boolean[n];
        int maxVal = 0;
        for (int v : arrivals) {
            maxVal = Math.max(maxVal, v);
        }
        int[] freq = new int[maxVal + 1];
        for (int i = 0; i < n; i++) {
            int outIdx = i - w;
            if (outIdx >= 0 && !removed[outIdx]) {
                int oldVal = arrivals[outIdx];
                freq[oldVal]--;
            }
            int val = arrivals[i];
            if (freq[val] >= m) {
                dis++;
                removed[i] = true;
            } else {
                freq[val]++;
            }
        }
        return dis;
    }
}

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