CF 1697B - Promo

Rating: 900
Tags: greedy, sortings
Solve time: 2m 6s
Verified: yes

Solution

Problem Understanding

We have a store with n items, each with a specific price. The store runs a promotion where, if a customer buys at least x items, the y cheapest among those items are free. For each query (x, y), we want to determine the maximum total value of items a customer can get for free.

The input gives the item prices as an array p of size n and then q independent queries. Each query specifies the promotion parameters (x, y), and the output is a single integer for each query: the maximum sum of the cheapest y items among any selection of x items.

The main constraints are that n and q can be up to 200,000, and item prices can go up to one million. A naive approach that iterates over all subsets of size x is impossible since the number of subsets grows combinatorially. Even iterating over all possible x-item selections explicitly would exceed 10^10 operations in the worst case. We need a solution that works in near-linear time with respect to n, ideally O(n log n) plus O(1) per query.

Non-obvious edge cases include when all items have the same price, when x equals n forcing the customer to buy everything, or when y equals x so every selected item is free. For example, if the prices are [5, 5, 5] and the query is (3, 3), the free value is 15. A careless approach that always tries to pick the globally cheapest items might miscompute the sum for other x values.

Approaches

The brute-force approach would be to generate all possible subsets of size x for a query, sort each subset, take the y cheapest items, and sum their prices. This works logically but is prohibitively expensive. Even for moderate n = 20, there are over 100,000 subsets to consider, and n = 2 * 10^5 is completely intractable.

The key observation is that to maximize the value of the y free items, we want to select the x most expensive items possible, because the y cheapest among them will still be higher in value than cheaper items outside the selection. Sorting the full array in descending order allows us to pick the first x items quickly. Then, among those x items, the y cheapest can be summed using a prefix sum array. This reduces the problem to a single sorting operation plus simple arithmetic per query.

This greedy insight is correct because the free items are always taken from the selected x, so maximizing the total sum of the free items is equivalent to taking the most expensive x and summing their y cheapest.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n choose x * x log x)	O(x) per query	Too slow
Optimal	O(n log n + q)	O(n)	Accepted

Algorithm Walkthrough

Read the number of items n and the number of queries q. Read the item prices into an array p.
Sort p in descending order. This ensures the first x elements of p for any query (x, y) are the most expensive selection.
Build a prefix sum array prefix where prefix[i] is the sum of the first i elements of p. This allows fast computation of sums over any contiguous segment.
For each query (x, y), the x most expensive items are the first x elements of p. Among these, the y cheapest are the last y elements in this segment. Compute their sum as prefix[x] - prefix[x-y].
Output the sum for each query.

Why it works: Sorting guarantees that any x-sized selection that maximizes the free items' sum is the first x elements. Prefix sums allow constant-time calculation of the y cheapest within that selection. The algorithm never selects a suboptimal set because moving any more expensive item outside the first x would reduce the sum of the free items.

Python Solution

import sys
input = sys.stdin.readline

n, q = map(int, input().split())
p = list(map(int, input().split()))

p.sort(reverse=True)

# prefix[i] = sum of first i elements
prefix = [0] * (n + 1)
for i in range(n):
    prefix[i + 1] = prefix[i] + p[i]

for _ in range(q):
    x, y = map(int, input().split())
    # sum of y cheapest in first x items
    print(prefix[x] - prefix[x - y])

The code first sorts the prices in descending order to ensure we can always pick the most expensive x items. The prefix sum array allows us to compute sums of contiguous segments in constant time. The subtle point is the indexing in prefix[x] - prefix[x - y]: prefix[x] sums the first x items, and subtracting prefix[x - y] removes the sum of the more expensive x-y items, leaving only the y cheapest.

Worked Examples

Sample Input 1

Step	Sorted Prices	Prefix	x	y	Sum
Query 1	[5,5,3,2,1]	[0,5,10,13,15,16]	3	2	13-5=8
Query 2	[5,5,3,2,1]	same	1	1	5-0=5
Query 3	[5,5,3,2,1]	same	5	3	16-10=6

This demonstrates that sorting plus prefix sums allows us to pick the x items and sum the y cheapest in O(1) per query.

Edge Case Example

Prices [1,1,1], query (3,3). Sorted prices [1,1,1], prefix [0,1,2,3]. Sum of last 3 of first 3 is 3-0=3. This confirms the algorithm handles y=x correctly.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n + q)	Sorting takes O(n log n), each query is O(1) using prefix sums
Space	O(n)	Prefix array of size n+1

For the constraints n, q ≤ 2 * 10^5, sorting dominates and fits comfortably within 2 seconds. Memory usage is linear in n.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    exec(open("solution.py").read())  # or paste solution here
    return output.getvalue().strip()

# Provided samples
assert run("5 3\n5 3 1 5 2\n3 2\n1 1\n5 3\n") == "8\n5\n6", "sample 1"

# Minimum size input
assert run("1 1\n10\n1 1\n") == "10", "min size"

# All equal values
assert run("4 2\n5 5 5 5\n2 1\n4 2\n") == "5\n10", "all equal"

# Maximum x
assert run("5 1\n1 2 3 4 5\n5 2\n") == "3", "x=n"

# Random large case
import random
n = 10
prices = " ".join(str(random.randint(1,10)) for _ in range(n))
assert run(f"{n} 1\n{prices}\n{n} 3\n")  # just ensure no crash

Test input	Expected output	What it validates
"1 1\n10\n1 1"	"10"	Minimum-size input
"4 2\n5 5 5 5\n2 1\n4 2"	"5\n10"	All equal prices, multiple queries
"5 1\n1 2 3 4 5\n5 2"	"3"	x equals n
Random 10-item case	varies	Large random input, checks no crash

Edge Cases

If all prices are equal, the algorithm still correctly sums the y cheapest, since prefix[x] - prefix[x-y] works regardless of equality. For y=x, the sum of all first x items is computed correctly. If y=1, it still picks the last of the first x items after sorting, ensuring the minimum of that segment. This handles all subtle boundaries without special-casing.