3679. Minimum Discards to Balance Inventory

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.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36

#include <iostream>
#include <vector>
#include <map>

class Solution {
public:
    int minArrivalsToDiscard(std::vector<int>& arrivals, int w, int m) {
        int num = 0;
        int n = arrivals.size();
        w = std::min(n, w);
        
        std::unordered_map<int, int> map;
        for (int i = 0; i < w; ++i) {
            if (map[arrivals[i]] < m) {
                ++map[arrivals[i]];
            }
            else {
                ++num;
                arrivals[i] = 0;
            }
        }
        
        for (int i = w; i < n; ++i) {
            --map[arrivals[i - w]];
            if (map[arrivals[i]] < m) {
                ++map[arrivals[i]];
            }
            else {
                ++num;
                arrivals[i] = 0;
            }
        }
        
        return num;
    }
};