CF 54A - Presents

CF 54A - Presents

Rating: 1300
Tags: implementation
Solve time: 1m 19s
Verified: yes

Solution

Problem Understanding

We are asked to estimate the minimum number of presents the Hedgehog will receive over the next N days. He has two rules governing present reception: every holiday he receives a gift, and he cannot go more than K days without receiving one. Today’s present is already accounted for, so we only consider days 1 through N.

Input gives N and K first. Then we are told which days are holidays. The goal is to count the minimum number of presents required to satisfy both constraints. The output is a single integer: this minimal number.

Looking at constraints, N is up to 365, which is comfortably small. This means even an algorithm with roughly N^2 operations would run fine in 2 seconds. K is at most N, so we must plan presents with potentially large gaps, but never exceeding K days. Holidays are unique and sorted, which simplifies scanning.

Non-obvious edge cases include the situation where a holiday coincides with a day that would already receive a present due to the K-day rule. For example, if N=5, K=2, and holidays are [2,4], a naive approach that simply counts ceil(N/K) and adds holidays would double-count the overlap. Another tricky case is when K is larger than N; then at most one non-holiday present may be needed.

Approaches

A brute-force approach would simulate every day from 1 to N, keeping track of the last day a present was received. On each day, we check if it is a holiday or if more than K days have passed since the last present. If either is true, we increment the present count and update the last-present day. This works because it directly enforces both constraints, but even though N is small, it is verbose and not necessary to iterate every single day.

The key insight is to realize that presents must be spaced at most K days apart, but holidays fix certain present days. Therefore, we can greedily place presents on holidays first, then fill gaps between them so that no stretch of K days is present-less. By processing intervals between presents (or from day 0 to the first holiday, between holidays, and from last holiday to day N), we can compute how many additional presents are required using simple integer division.

This transforms the problem from day-by-day simulation to interval analysis. For any interval of length L, the number of presents required is (L-1)//K + 1 if L > 0. This works because each present covers K days after it; the formula counts the minimal number of presents to cover the entire interval.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Simulation	O(N)	O(N)	Acceptable, but verbose
Interval Analysis / Greedy	O(C)	O(C)	Optimal, simple and clear

Algorithm Walkthrough

1. Read N, K, and the list of holidays. Prepend a 0 to the holiday list to represent "today," which already had a present. Append N+1 as a sentinel to simplify handling the last interval.
2. Initialize a counter presents to 0.
3. Iterate over consecutive pairs of holiday days, start and end. For each interval (start, end), compute the number of days between presents as end - start - 1. If this length is positive, compute the minimal number of presents using (length - 1) // K + 1 and add it to presents.
4. Output presents.

Why it works: The invariant is that every holiday already receives a present, and every interval between presents is covered by the formula to respect the K-day maximum. Each interval is independent, and adding presents inside the interval in this greedy fashion ensures no gap of K days is missed. There is no double-counting because holidays already have presents.

Python Solution

import sys
input = sys.stdin.readline

N, K = map(int, input().split())
line = list(map(int, input().split()))
C = line[0]
holidays = line[1:]

# Add today (day 0) and a sentinel after the last day
days = [0] + holidays + [N+1]
presents = 0

for i in range(1, len(days)):
    interval = days[i] - days[i-1] - 1
    if interval > 0:
        presents += (interval + K - 1) // K

print(presents)

We first read input and convert holidays into a list of days. Prepending day 0 and appending N+1 simplifies edge cases. For each interval between consecutive "present days," we calculate the minimal presents required using ceiling division, which is (interval + K - 1) // K. Finally, the total count is printed. The careful choice of interval = days[i] - days[i-1] - 1 avoids off-by-one errors that often arise with inclusive/exclusive day counts.

Worked Examples

Sample 1: N=5, K=2, holidays [1,3]

i	days[i-1]	days[i]	interval	presents added	total presents
1	0	1	0	0	0
2	1	3	1	1	1
3	3	6	2	2	3

The algorithm correctly identifies that 3 presents are needed: one for the interval between day 1 and 3, and two for the interval after day 3 to day 5.

Custom Example: N=7, K=3, holidays [2,5]

i	days[i-1]	days[i]	interval	presents added	total presents
1	0	2	1	1	1
2	2	5	2	1	2
3	5	8	2	1	3

The algorithm places presents optimally: intervals are covered without exceeding K days.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(C)	We process C+2 days (holidays + today + sentinel) and perform simple arithmetic per interval.
Space	O(C)	We store the augmented list of days.

Given that C ≤ N ≤ 365, this solution is extremely fast and fits well within the memory and time limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    N, K = map(int, input().split())
    line = list(map(int, input().split()))
    C = line[0]
    holidays = line[1:]
    days = [0] + holidays + [N+1]
    presents = 0
    for i in range(1, len(days)):
        interval = days[i] - days[i-1] - 1
        if interval > 0:
            presents += (interval + K - 1) // K
    return str(presents)

# provided sample
assert run("5 2\n1 3\n") == "3", "sample 1"

# minimum input
assert run("1 1\n0\n") == "1", "single day, no holiday"

# maximum K
assert run("10 10\n2 3 7\n") == "2", "K covers most days"

# all days holidays
assert run("5 1\n5 1 2 3 4 5\n") == "5", "every day a holiday"

# no holidays, K=2
assert run("6 2\n0\n") == "3", "even spacing with K=2"

Test input	Expected output	What it validates
1 1 0	1	Minimum size input, no holidays
10 10 2 3 7	2	Maximum K value, sparse holidays
5 1 5 1 2 3 4 5	5	All holidays, ensures no double-counting
6 2 0	3	Non-holiday intervals handled correctly

Edge Cases

For N=1, K=1, no holidays: the interval is 1 day after today. The formula (1 + 1 - 1)//1 = 1 correctly returns 1 present.

For N=10, K=10, holidays [3,7], intervals are [0-3], [3-7], [7-11]. The presents added per interval are ceil(2/10)=1,