CF 1995B1 - Bouquet (Easy Version)
Each flower has two properties that are actually the same number: its petal count and its price. A flower with k petals costs exactly k coins. We may buy any subset of the flowers available in the store.
CF 1995B1 - Bouquet (Easy Version)
Rating: 1100
Tags: binary search, brute force, greedy, sortings, two pointers
Solve time: 2m 19s
Verified: yes
Solution
Problem Understanding
Each flower has two properties that are actually the same number: its petal count and its price. A flower with k petals costs exactly k coins.
We may buy any subset of the flowers available in the store. The bouquet is valid only if the difference between the petal counts of any two chosen flowers is at most one. Since all flowers are individual objects in this easy version, each flower can either be taken or not taken.
The goal is to maximize the total number of petals while spending at most m coins.
A useful observation appears immediately: because cost equals petals for every flower, maximizing petals under a budget is exactly the same as maximizing total cost while keeping the total cost at most m.
The constraint on petal differences is what makes the problem interesting. If the bouquet contains a flower with x petals, then every other flower must have either x or x + 1 petals. Any larger gap would violate the condition.
The sum of all n across test cases is at most 2 · 10^5. That means an O(n^2) solution would require roughly 4 · 10^10 operations in the worst case, which is far beyond the limit. Sorting each test case and then doing linear processing is completely feasible because O(n log n) over 2 · 10^5 elements is only a few million operations.
Several edge cases can easily break a careless solution.
Consider:
n = 3, m = 10
flowers = [2, 2, 2]
The answer is 6. A solution that always tries to spend the entire budget would incorrectly look for total cost 10, even though only 6 petals are available.
Consider:
n = 3, m = 7
flowers = [2, 2, 3]
The answer is 7, using all flowers. A solution that only considers flowers with exactly one petal value would miss the optimal bouquet.
Consider:
n = 4, m = 8
flowers = [1, 3, 3, 3]
The answer is 6, obtained from two flowers with 3 petals. Taking the flower with 1 petal together with any 3-petal flower is illegal because the difference is 2.
These examples show that we must simultaneously enforce both the budget constraint and the petal-difference constraint.
Approaches
A brute-force solution would enumerate every subset of flowers, compute its total cost, check whether the petal values differ by at most one, and keep the best valid answer.
This works because it directly tests the definition of a valid bouquet. Unfortunately, there are 2^n subsets. Even for n = 50 this is hopeless, while the actual limit is 2 · 10^5.
The key observation is that after sorting the flowers by petal count, every valid bouquet corresponds to a contiguous segment whose minimum and maximum values differ by at most one.
Why does sorting help? Suppose we sort the petal counts. If some chosen flowers contain only values x and x + 1, then all such flowers appear together in the sorted order. Any flower outside that region would have a value either smaller than x or larger than x + 1.
Now think about the objective. Since every flower has positive cost, if a segment is valid but its total cost exceeds m, we can only improve feasibility by removing flowers from the left side. This is exactly the setting where a sliding window, also called two pointers, excels.
We maintain a window [l, r] in the sorted array. The window must satisfy two conditions:
First, the largest and smallest values differ by at most one.
Second, the sum of values in the window does not exceed m.
Whenever one of these conditions is violated, we move the left pointer forward until both become true again. Every valid window represents a feasible bouquet, and its total petal count equals its sum.
The maximum valid window sum is the answer.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | O(2^n · n) | O(n) | Too slow |
| Optimal Two Pointers | O(n log n) | O(n) | Accepted |
Algorithm Walkthrough
- Sort the petal counts.
- Initialize two pointers
l = 0andr = 0, along with a running window sum. - Extend the window by moving
rfrom left to right. Adda[r]to the running sum. - While the window is invalid, move
lforward and subtracta[l]from the running sum. - A window is invalid if either:
a[r] - a[l] > 1, meaning the bouquet contains petal counts differing by more than one.
The running sum exceeds m, meaning the bouquet costs too much.
6. Once the window becomes valid again, update the answer with the current window sum.
7. Continue until every position has been used as the right endpoint.
8. Output the largest valid sum found.
Why it works
At every moment, the window contains a contiguous range of the sorted array. The first condition guarantees that all flowers inside the window have petal counts differing by at most one. The second condition guarantees that the bouquet fits within the budget.
When the window becomes invalid, removing flowers from the left is the only useful action. Because all flower costs are positive, enlarging the window further cannot repair an excessive sum. Similarly, if the petal range is too large, removing the smallest values is exactly what shrinks that range.
Every valid contiguous segment appears as the window during the sweep, and the algorithm records the sum of every feasible segment. Since the answer is the maximum feasible sum, the algorithm is correct.
Python Solution
import sys
input = sys.stdin.readline
def solve():
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
l = 0
cur_sum = 0
ans = 0
for r in range(n):
cur_sum += a[r]
while l <= r and (a[r] - a[l] > 1 or cur_sum > m):
cur_sum -= a[l]
l += 1
ans = max(ans, cur_sum)
print(ans)
if __name__ == "__main__":
solve()
The first step sorts the flowers. After sorting, every valid bouquet using only values x and x + 1 becomes a contiguous segment.
cur_sum stores the total petals, which is also the total cost of the current window. This allows budget checking in constant time.
The while loop is the heart of the solution. It keeps shrinking the left side until both constraints hold simultaneously. Using a loop rather than a single if is essential because one removal may not be enough to restore validity.
The answer is updated only after the window is valid. Since every valid window is considered exactly once, the maximum recorded sum is the optimal bouquet value.
Python integers automatically handle values up to 10^18, so no special overflow handling is required.
Worked Examples
Example 1
Input:
n = 5, m = 10
flowers = [1, 1, 2, 2, 3]
After sorting:
[1, 1, 2, 2, 3]
| r | Value | Window | Sum | Valid? | Answer |
|---|---|---|---|---|---|
| 0 | 1 | [1] | 1 | Yes | 1 |
| 1 | 1 | [1,1] | 2 | Yes | 2 |
| 2 | 2 | [1,1,2] | 4 | Yes | 4 |
| 3 | 2 | [1,1,2,2] | 6 | Yes | 6 |
| 4 | 3 | [1,1,2,2,3] | 9 | No, range=2 | 6 |
Shrink from left:
[1,2,2,3] sum=8, range=2
[2,2,3] sum=7, range=1
| r | Value | Window | Sum | Valid? | Answer |
|---|---|---|---|---|---|
| 4 | 3 | [2,2,3] | 7 | Yes | 7 |
Final answer: 7.
This example demonstrates the petal-difference constraint. Even though the total sum 9 fits the budget, the window is invalid because it contains both 1 and 3.
Example 2
Input:
n = 8, m = 20
flowers = [4, 2, 7, 5, 6, 1, 1, 1]
Sorted:
[1,1,1,2,4,5,6,7]
| r | Value | Sum Before Shrink | Action |
|---|---|---|---|
| 0 | 1 | 1 | keep |
| 1 | 1 | 2 | keep |
| 2 | 1 | 3 | keep |
| 3 | 2 | 5 | keep |
| 4 | 4 | 9 | shrink until range ≤ 1 |
| 5 | 5 | 9 | keep |
| 6 | 6 | 15 | shrink until range ≤ 1 |
| 7 | 7 | 18 | shrink until range ≤ 1 |
The best valid window encountered is [6,7] with sum 13.
Final answer: 13.
This trace shows that sorting naturally groups compatible petal counts together. Whenever a gap larger than one appears, the left pointer advances until only acceptable values remain.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n log n) | Sorting dominates, the two pointers sweep is O(n) |
| Space | O(n) | Storage for the flower array |
The total number of flowers across all test cases is at most 2 · 10^5. Sorting that many values and performing a linear scan easily fits within the 2-second limit and the memory limit.
Test Cases
# helper: run solution on input string, return output string
import sys
import io
def run(inp: str) -> str:
sys.stdin = io.StringIO(inp)
input = sys.stdin.readline
t = int(input())
out = []
for _ in range(t):
n, m = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
l = 0
cur = 0
ans = 0
for r in range(n):
cur += a[r]
while l <= r and (a[r] - a[l] > 1 or cur > m):
cur -= a[l]
l += 1
ans = max(ans, cur)
out.append(str(ans))
return "\n".join(out)
# provided sample
assert run(
"""5
5 10
1 1 2 2 3
8 20
4 2 7 5 6 1 1 1
8 100000
239 30 610 122 24 40 8 2
11 13
2 4 11 1 1 2 3 5 4 3 2
8 1033
206 206 206 207 207 207 207 1000
"""
) == """7
13
610
13
1033"""
# minimum size
assert run(
"""1
1 5
3
"""
) == "3"
# all equal values
assert run(
"""1
5 20
4 4 4 4 4
"""
) == "20"
# budget forces shrinking
assert run(
"""1
3 7
2 2 3
"""
) == "7"
# gap larger than one
assert run(
"""1
4 8
1 3 3 3
"""
) == "6"
| Test input | Expected output | What it validates |
|---|---|---|
1 flower worth 3, budget 5 |
3 |
Minimum-size instance |
4 4 4 4 4, budget 20 |
20 |
All-equal values |
2 2 3, budget 7 |
7 |
Exact budget match |
1 3 3 3, budget 8 |
6 |
Difference constraint enforcement |
Edge Cases
Consider:
1
3 10
2 2 2
The entire array forms a valid window. The running sum becomes 6, which never exceeds the budget. The algorithm returns 6, correctly handling the case where the budget is larger than the total value of all available flowers.
Consider:
1
3 7
2 2 3
The sorted array is already [2,2,3]. The window expands to contain all three flowers, producing sum 7. The range is 1, so the bouquet is valid. The algorithm returns 7, correctly allowing two adjacent petal values.
Consider:
1
4 8
1 3 3 3
After sorting, the first time a 3 enters the window the range becomes 2. The shrinking loop removes the 1, leaving only 3s. The best feasible sum is 6, which the algorithm records. This confirms that flowers differing by more than one petal are never kept together.
Consider:
1
4 5
2 2 2 2
The window sum eventually reaches 8, exceeding the budget. The shrinking loop repeatedly removes flowers until the sum becomes 4. The algorithm returns 4, showing that the budget constraint is enforced independently of the petal-difference constraint.