Package g3601_3700.s3679_minimum_discards_to_balance_inventory

Class Solution

java.lang.Object

g3601_3700.s3679_minimum_discards_to_balance_inventory.Solution

public class Solution
extends Object

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

Constructor Summary

Constructors

Constructor	Description
Solution()

Method Summary

Modifier and Type	Method	Description
int	minArrivalsToDiscard(int[] arrivals, int w, int m)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Details

Solution

public Solution()

Method Details

minArrivalsToDiscard

public int minArrivalsToDiscard(int[] arrivals,
 int w,
 int m)