CF 104879A - Coffee Cocktail

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Solve time: 46s
Verified: yes

Solution

Problem Understanding

We are given several kinds of snacks, where each snack type can appear multiple times and each individual snack contributes some amount of caffeine. The key simplification is that instead of thinking in terms of raw mass and percentage, we directly treat each snack as contributing a fixed caffeine value. So every item in the input can be viewed as a “packet” that adds a certain amount of caffeine.

The task is to reach at least a required caffeine threshold using a minimum number of chosen snack types. The important detail is that selecting one snack of a type effectively allows us to use all snacks of that same type for the purpose of accumulating caffeine, so the decision is made at the level of types rather than individual items.

The output is the minimum number of distinct snack types we need to take so that the total caffeine from all selected snacks across those types reaches at least the target value, or −1 if even taking all snacks is insufficient.

The constraints imply that the input size can be large, so any solution that tries all subsets of types or recomputes sums repeatedly would be too slow. A solution with sorting or linear passes per test is expected, so roughly O(n log n) or O(n) per test is acceptable, while exponential or quadratic behavior is not.

A subtle failure case arises when treating snacks independently rather than grouped by type. If we greedily pick individual snacks without aggregating by type, we may select multiple snacks from a low-yield type when it would have been better to fully commit to a higher-yield type earlier.

For example, suppose we have a type A with snacks contributing 100 total caffeine across many items, and a type B with two snacks contributing 51 each, totaling 102. If we pick greedily by individual values, we might take both 51s first and delay picking type A, but the correct reasoning depends on grouping by type contribution, not per-item selection.

Approaches

The brute-force perspective starts by considering every possible subset of snack types. For each subset, we sum all caffeine contributions from all snacks belonging to those types and check whether the total reaches the target. Among all valid subsets, we pick the one with the smallest number of types. This is correct because it directly evaluates every possible decision. However, if there are q types, this requires examining 2^q subsets, and computing sums for each subset can take linear time in the number of items. This quickly becomes infeasible even for moderately sized inputs.

The key observation is that within each type, there is no reason to partially use its contribution. Once we decide to include a type, we take all of its snacks, since each contributes positively to the total and there is no penalty for doing so. This collapses the problem from item-level decisions to type-level weights, where each type has a single aggregate value equal to its total caffeine.

Once reduced to type-level weights, the goal becomes selecting the minimum number of types whose total weight reaches at least x. This is a classic greedy structure: we want to accumulate as much caffeine as possible using as few types as possible, so we always prefer the most “valuable” type first.

This reduces the task to computing total caffeine per type, sorting types by this total in descending order, and then greedily accumulating until we reach the target.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force over type subsets	O(2^q · n)	O(n)	Too slow
Group + sort greedy	O(n log n)	O(n)	Accepted

Algorithm Walkthrough

Read all snacks and group them by type, accumulating the total caffeine contribution for each type. This step transforms item-level data into a compact representation where each type has a single meaningful weight.
For each type, compute its total caffeine contribution. This is necessary because only the aggregate matters for any decision about selecting that type.
Store all type totals in a list and sort them in descending order. Sorting ensures we always consider the most impactful type first, which is essential for minimizing the number of types chosen.
Initialize a running sum at zero and a counter for selected types.
Iterate through the sorted type totals, adding each one to the running sum and incrementing the type counter. After each addition, check whether the running sum has reached or exceeded the target caffeine value.
If the target is reached, output the number of types used so far. If the loop ends without reaching the target, output −1.

The reason sorting is safe is that every type contributes independently and fully once selected, so rearranging selection order cannot improve the number of types needed beyond always taking the largest remaining contribution first.

Why it works

At any point in the process, the only relevant state is how much caffeine has been accumulated and how many types have been chosen. Since each type contributes a fixed positive amount, replacing a smaller contribution with a larger one earlier can never increase the number of types required to reach the same total. This establishes that an optimal solution always exists that selects types in non-increasing order of total caffeine.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n, x = map(int, input().split())
    
    from collections import defaultdict
    tot = defaultdict(int)

    for _ in range(n):
        t, c = map(int, input().split())
        tot[t] += c

    vals = sorted(tot.values(), reverse=True)

    cur = 0
    cnt = 0

    for v in vals:
        cur += v
        cnt += 1
        if cur >= x:
            print(cnt)
            return

    print(-1)

if __name__ == "__main__":
    solve()

The implementation mirrors the algorithm directly. The dictionary aggregates caffeine per type in a single pass, ensuring linear complexity in the number of snacks. Sorting the aggregated values ensures we always pick the best available type next.

The only subtle point is that the grouping step must happen before sorting. Sorting raw items instead of type aggregates would lead to incorrect greedy decisions, since multiple items from the same type could be split across the selection unnecessarily.

Worked Examples

Consider an input with three types:

Input:

Here type 1 totals 80, type 2 totals 60, and type 3 totals 60.

Step	Chosen types	Current sum	Action
1	[]	0	Start
2	[1]	80	Take largest type
3	[1, 2]	140	Add next largest
4	[1, 2, 3]	200	Target reached

We stop after selecting 3 types.

This trace shows that even though type 2 and 3 are equal, the ordering between them does not matter. What matters is that we always prioritize larger contributions first.

Now consider a case where reaching the target is impossible:

Input:

Totals are 50, 60, 40. Even after selecting all types, we only reach 150, so the answer is −1.

Step	Chosen types	Current sum	Action
1	[2]	60	Take largest
2	[2,1]	110	Continue
3	[2,1,3]	150	Exhaust all
4	-	150	Fail

This confirms that the algorithm correctly handles infeasible targets.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	grouping is O(n), sorting type totals dominates
Space	O(n)	storing totals per type

The solution fits comfortably within typical constraints for Codeforces problems with up to 2e5 items, since sorting dominates and everything else is linear.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    return sys.stdout.getvalue() if False else solve_capture(inp)

def solve_capture(inp: str) -> str:
    import sys
    from collections import defaultdict
    input = sys.stdin.readline

    it = iter(inp.strip().split())
    n = int(next(it))
    x = int(next(it))

    tot = defaultdict(int)
    for _ in range(n):
        t = int(next(it))
        c = int(next(it))
        tot[t] += c

    vals = sorted(tot.values(), reverse=True)

    cur = 0
    cnt = 0

    for v in vals:
        cur += v
        cnt += 1
        if cur >= x:
            return str(cnt)

    return str(-1)

# sample-like cases
assert solve_capture("5 150 1 50 1 30 2 60 3 40 3 20") == "3"
assert solve_capture("3 200 1 50 2 60 3 40") == "-1"

# minimal case
assert solve_capture("1 10 1 10") == "1"

# already satisfied by one type
assert solve_capture("4 50 1 10 1 20 1 30 2 100") == "1"

# multiple small types needed
assert solve_capture("6 100 1 20 2 20 3 20 4 20 5 20 6 20") == "5"

Test input	Expected output	What it validates
single exact match	1	minimal boundary case
impossible sum	-1	feasibility check
one dominant type	1	greedy early stop correctness
many equal small types	5	ordering stability and accumulation

Edge Cases

A first edge case is when a single type already exceeds the target. The algorithm handles this immediately because after sorting, the largest type is processed first and triggers termination in one step.

Another case is when all types have very small contributions, and the answer depends on combining many of them. Since the algorithm accumulates in sorted order, it naturally keeps adding until the threshold is reached or exhaustion occurs, without requiring any special handling.

A final case is when no combination reaches the target. Because the loop finishes after processing all aggregated type values, and the sum never crosses the threshold, the algorithm correctly returns −1 without premature termination.