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Harbor Banner Schedule Slots

Published at05 Jan 2026
Created by
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You are given the travel days that need coverage and the prices of different pass lengths. The goal is to cover every required day with the minimum total cost.

A pass does not have to start at day 1. You buy it on a specific travel day, and it covers a block of future days depending on its duration. That means sometimes it is cheaper to buy one longer pass, and sometimes several shorter passes work better.

For example, if days = [1,2,3,4,5,6,7] and costs = [3,8,20], the answer is 9 because three 3-day passes are cheaper than other combinations. If there are no travel days, the answer is 0 because no pass is needed at all.

So the task is to choose passes that cover all required days while keeping the total cost as low as possible.

Example Input & Output

Example 1
Input
days = [1,2,3,4,5,6,7], costs = [3,8,20]
Output
9
Explanation

Three 3-day passes cover the entire week for cost 9.

Example 2
Input
days = [], costs = [5,10,20]
Output
0
Explanation

No scheduled nights mean no passes are needed.

Example 3
Input
days = [1,4,6,7,8,20], costs = [2,7,15]
Output
11
Explanation

Buy 1-day passes for nights 1,4,6 and a 7-day pass covering 7,8,20.

Algorithm Flow

Recommendation Algorithm Flow for Harbor Banner Schedule Slots
Recommendation Algorithm Flow for Harbor Banner Schedule Slots

Solution Approach

This is a dynamic programming problem over the travel-day list. Let dp[i] be the minimum cost needed to cover travel days starting from index i.

At each travel day, you try buying each available pass type. A pass covers the current day and also skips forward past every future travel day that falls inside its coverage window.

So the transition looks like this idea:

dp[i] = min(
  cost1 + dp[nextIndexAfterPass1],
  cost2 + dp[nextIndexAfterPass2],
  cost3 + dp[nextIndexAfterPass3]
)

The base case is that once all travel days are covered, the remaining cost is 0.

You can implement this with memoized recursion or bottom-up DP. The key is that every decision at day i only depends on which pass you buy there and which travel day is the first one not covered by that pass.

Best Answers

java
import java.util.*;
class Solution {
    public int calculate_min_slots(int[][] intervals) {
        int[] starts = new int[intervals.length];
        int[] ends = new int[intervals.length];
        for (int i = 0; i < intervals.length; i++) {
            starts[i] = intervals[i][0];
            ends[i] = intervals[i][1];
        }
        Arrays.sort(starts);
        Arrays.sort(ends);
        int slots = 0, pEnd = 0;
        for (int s : starts) {
            if (s < ends[pEnd]) slots++;
            else pEnd++;
        }
        return slots;
    }
}