CF 1982C - Boring Day

We are asked to simulate a game with a deck of cards arranged in a specific order. Each card has a positive integer on it.

codeforcescompetitive-programmingbinary-searchdata-structuresdpgreedytwo-pointers

CF 1982C - Boring Day

Rating: 1200
Tags: binary search, data structures, dp, greedy, two pointers
Solve time: 1m 55s
Verified: yes

Solution

Problem Understanding

We are asked to simulate a game with a deck of cards arranged in a specific order. Each card has a positive integer on it. Egor takes a non-zero number of cards from the top of the deck each round, and the round is won if the sum of the taken cards lies within a given range $[l, r]$. The task is to maximize the total number of rounds won, but there is no requirement that the winning rounds occur consecutively. Input gives multiple independent test cases, each specifying the number of cards, the target range, and the sequence of card values. Output for each test case is a single integer: the maximum number of winning rounds.

Constraints allow up to $10^5$ cards per test case, and the sum of $n$ over all test cases is $2 \cdot 10^5$. This rules out $O(n^2)$ solutions, since iterating over all possible contiguous subarrays for each round would result in over $10^{10}$ operations. Each card value and the bounds $l, r$ can be up to $10^9$, so we must carefully handle integer sums without relying on array indexing tricks that assume small numbers.

Edge cases include sequences where all cards are too large to ever fit in $[l, r]$ individually, sequences where individual cards already satisfy the range, and sequences where multiple small cards must be grouped to reach the range. For example, with cards [1, 2, 1] and range [3, 3], we cannot take any single card, but taking the first two cards sums to 3, yielding a winning round. A naive greedy approach of always taking the top card and checking if it wins may miss opportunities to combine small cards.

Approaches

The brute-force approach is to examine every possible prefix of the remaining deck to find a contiguous sum in the winning range. While this is correct, it requires $O(n^2)$ per test case because, in the worst case, we might have to check all possible prefix lengths at each step. With $n$ up to $10^5$, this is too slow.

The key observation is that the problem reduces to counting the number of prefixes whose sums fall in $[l, r]$. Consider the prefix sums array $S$ where $S_i = a_1 + a_2 + ... + a_i$. Each round corresponds to choosing a starting and ending index of a prefix such that the sum is within $[l, r]$. Since Egor must always take cards from the top, each round starts at the first remaining card. Therefore, we only need to find, for each position, the shortest prefix starting from the current card that lies in $[l, r]$. The sum of the first $k$ cards is $S_k$, so the problem becomes finding the largest $k$ such that $S_k - S_{prev}$ lies in $[l, r]$, where $S_{prev}$ is the sum of all previously removed cards. Using a two-pointer technique on prefix sums allows us to compute this in linear time.

Another insight is that since card values are positive, sums are strictly increasing as we include more cards. This ensures that the two-pointer approach will always find the minimal-length prefix satisfying the range without backtracking, as any larger prefix is also strictly larger and can be checked sequentially.

Approach Time Complexity Space Complexity Verdict
Brute Force O(n²) O(n) Too slow
Two-Pointer on Prefix Sums O(n) O(n) Accepted

Algorithm Walkthrough

  1. For each test case, read $n$, $l$, $r$, and the array of card values $a$. Initialize a prefix sum array $S$ with $S_0 = 0$ and $S_i = S_{i-1} + a_i$. This allows us to compute the sum of any top segment in $O(1)$ as $S[j] - S[i-1]$.
  2. Initialize a counter wins = 0 and a pointer start = 0, which indicates the position of the first card remaining in the deck.
  3. Use a pointer end to iterate over the prefix sums from start + 1 up to n. For each end, compute sum_segment = S[end] - S[start].
  4. If sum_segment exceeds r, we cannot take more cards for this round, so increment start until sum_segment <= r or start = end. This is valid because all card values are positive; adding more cards only increases the sum.
  5. If sum_segment is at least l and at most r, it counts as a winning round. Increment wins and move start to end to begin the next round with the next card.
  6. Repeat until all cards are processed. Output wins for this test case.

Why it works: The algorithm maintains the invariant that start always points to the first uncollected card, and end grows monotonically. Since card values are positive, the sum from start to end increases with end. Two pointers ensure we skip over segments that are too small or too large efficiently. Each card is visited at most twice (once by start and once by end), guaranteeing linear time per test case.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n, l, r = map(int, input().split())
        a = list(map(int, input().split()))
        
        wins = 0
        start = 0
        prefix_sum = [0]*(n+1)
        for i in range(n):
            prefix_sum[i+1] = prefix_sum[i] + a[i]
        
        end = 1
        while start < n:
            while end <= n and prefix_sum[end] - prefix_sum[start] < l:
                end += 1
            if end <= n and prefix_sum[end] - prefix_sum[start] <= r:
                wins += 1
                start = end
                end = start + 1
            else:
                start += 1
                if end <= start:
                    end = start + 1
        print(wins)

if __name__ == "__main__":
    solve()

The code initializes prefix sums to quickly compute segment sums. start marks the first uncollected card, while end searches for the shortest segment whose sum is in the winning range. We increment wins when a segment fits the criteria and move start to end to start the next round. The nested loops never backtrack because all values are positive, so each index is processed linearly.

Worked Examples

Sample 1:

Input: 5 3 10 with cards [2, 1, 11, 3, 7]

start end sum_segment wins
0 1 2 0
0 2 3 1 (take first two cards)
start → 2 end = 3 11 1
start → 3 end = 4 3 2 (take 3)
start → 4 end = 5 7 3 (take 7)

Sample 2:

Input: 10 1 5 with cards [17, 8, 12, 11, 7, 11, 21, 13, 10, 8]

start end sum_segment wins
0 1 17 too large
start → 1 end = 2 8 too large
... ... ... 0

No segment fits in [1,5], so wins = 0.

These traces demonstrate that the two-pointer method correctly identifies winning rounds and skips unfit segments without missing combinations.

Complexity Analysis

Measure Complexity Explanation
Time O(n) per test case Each card is visited at most twice by start and end pointers.
Space O(n) Prefix sum array of size n+1

The total sum of n across all test cases is $2 \cdot 10^5$, making the total operations under $10^6$. Memory usage is also within the 256 MB limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    sys.stdout = io.StringIO()
    solve()
    return sys.stdout.getvalue().strip()

# provided samples
assert run("8\n5 3 10\n2 1 11 3 7\n10 1 5\n17 8 12 11 7 11 21 13 10 8\n3 4 5\n3 4 2\n8 12 25\n10 7 5 13 8 9 12 7\n2 3 3\n5 2\n9 7 9\n2 10 5 1 3 7 6 2 3\n1