Harbor Banner Schedule Slots

Like12

Harbor Banner Schedule Slots is an interval-merging problem even if the story sounds like planning crews You are given a list of time slots represented as start-end pairs and the goal is to merge all overlapping or touching slots so the final schedule has non-overlapping ranges The output should preserve chronological order after merging and include each maximal merged interval exactly once A standard strategy is sort by start time then scan from left to right Keep a current interval if the next interval starts before or at the current end merge by extending the end to the larger value Otherwise push the current interval to results and start a new one This approach is reliable and easy to reason about especially when input order is unsorted and overlaps chain across multiple adjacent entries The judge checks empty input already disjoint intervals full containment cases and edge-touching boundaries like 1 5 with 5 10 that should become one merged block in this dataset Return only the merged interval list structure expected by the platform with no extra logs This hard label comes from careful boundary handling and ensuring merge semantics stay consistent across all interval arrangements Be careful with interval semantics in this challenge touching boundaries are treated as mergeable so equality at endpoints should still combine ranges That one detail is easy to miss and is often the reason otherwise-correct interval solutions fail hidden tests on boundary-heavy schedules When

Examples

Example 1

Input

days = [1,2,3,4,5,6,7], costs = [3,8,20]

Output

Explanation

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

Example 2

Input

days = [], costs = [5,10,20]

Output

Explanation

No scheduled nights mean no passes are needed.

Example 3

Input

days = [1,4,6,7,8,20], costs = [2,7,15]

Output

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 - Budibadu

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;
    }
}

javascript

function calculate_min_slots(intervals) {
    const starts = intervals.map(i => i[0]).sort((a, b) => a - b);
    const ends = intervals.map(i => i[1]).sort((a, b) => a - b);
    let slots = 0, pEnd = 0;
    for (let s of starts) {
        if (s < ends[pEnd]) slots++;
        else pEnd++;
    }
    return slots;
}

php

function calculate_min_slots($intervals) {
    $starts = array_map(function($i) { return $i[0]; }, $intervals);
    $ends = array_map(function($i) { return $i[1]; }, $intervals);
    sort($starts);
    sort($ends);
    $slots = 0; $pEnd = 0;
    foreach ($starts as $s) {
        if ($s < $ends[$pEnd]) $slots++;
        else $pEnd++;
    }
    return $slots;
}

python

def calculate_min_slots(intervals: list) -> int:
    starts = sorted([i[0] for i in intervals])
    ends = sorted([i[1] for i in intervals])
    res = 0
    p_end = 0
    for s in starts:
        if s < ends[p_end]:
            res += 1
        else:
            p_end += 1
    return res

rust

pub fn calculate_min_slots(intervals: Vec<Vec<i32>>) -> i32 {
    let mut starts: Vec<i32> = intervals.iter().map(|i| i[0]).collect();
    let mut ends: Vec<i32> = intervals.iter().map(|i| i[1]).collect();
    starts.sort();
    ends.sort();
    let mut slots = 0;
    let mut p_end = 0;
    for s in starts {
        if s < ends[p_end] { slots += 1; }
        else { p_end += 1; }
    }
    slots
}

typescript

function calculate_min_slots(intervals: number[][]): number {
    const starts = intervals.map(i => i[0]).sort((a, b) => a - b);
    const ends = intervals.map(i => i[1]).sort((a, b) => a - b);
    let slots = 0, pEnd = 0;
    for (let s of starts) {
        if (s < ends[pEnd]) slots++;
        else pEnd++;
    }
    return slots;
}

Comments (0)

Join the Discussion

Share your thoughts, ask questions, or help others with this problem.

Harbor Banner Schedule Slots

Examples

Algorithm Flow

Best Answers

Solutions Coming Soon

Related Dynamic Programming Problems

Garden Lantern Seating Ways

Festival Glow Step Combinations

Evening Harmony Partition Count

Twilight Lantern Garden

Lantern Vowel Pathways

Celestial Pattern Synthesis

Comments (0)

Join the Discussion