CF 169A - Chores

CF 169A - Chores

Rating: 800
Tags: sortings
Solve time: 1m 25s
Verified: yes

Solution

Problem Understanding

We have a list of chore difficulties. Vasya must receive exactly b chores whose difficulty is at most x, while Petya must receive exactly a chores whose difficulty is strictly greater than x.

The value x must be an integer. We need to count how many integers satisfy the split.

After sorting the difficulties, the condition becomes much easier to reason about. Since exactly b chores must satisfy h[i] ≤ x, the first b smallest chores must belong to Vasya, and the remaining a chores must belong to Petya.

That immediately creates two boundary values:

• The largest chore Vasya gets is the b-th smallest value.
• The smallest chore Petya gets is the (b + 1)-th smallest value.

If we denote them as:

• L = sorted_h[b - 1]
• R = sorted_h[b]

then x must satisfy:

• x ≥ L, otherwise Vasya would receive fewer than b chores.
• x < R, otherwise Petya would receive fewer than a chores.

So every valid integer x lies in the interval [L, R - 1].

The constraints are tiny here, only up to 2000 elements, so even quadratic solutions would pass comfortably. Still, the structure of the problem leads to a very compact O(n log n) solution after sorting.

The tricky cases come from duplicate values around the boundary.

Consider this input:

7 3 4
1 2 3 3 3 4 5

After sorting, the fourth smallest value is 3, and the fifth smallest value is also 3.

We would need:

3 ≤ x < 3

which is impossible. The correct answer is 0.

A careless implementation might try to count numbers between the two boundary positions inclusively and accidentally return 1.

Another subtle case happens when there is a large gap:

5 2 3
1 2 10 20 30

Here:

L = 2
R = 10

Any integer from 2 through 9 works, so the answer is 8.

The valid values are not restricted to existing difficulties. They are all integers in the gap.

Approaches

The brute-force idea is straightforward. We can try every possible integer x, count how many chores satisfy h[i] ≤ x, and check whether that count equals b.

The smallest difficulty is at least 1, and the largest can reach 10^9, so iterating through all possible x values is completely unrealistic. Even if each check took only O(n), scanning up to 10^9 integers would require around 2 × 10^12 operations in the worst case.

The key observation is that only the boundary between Vasya's chores and Petya's chores matters.

Once the array is sorted, the first b elements must belong to Vasya and the remaining a elements must belong to Petya. The exact values inside those groups do not matter anymore.

Suppose:

sorted_h[b - 1] = L
sorted_h[b] = R

For Vasya to still receive exactly b chores, x must be at least L, because Vasya's hardest assigned chore must satisfy h[i] ≤ x.

For Petya to still receive exactly a chores, x must be strictly smaller than R, because Petya's easiest assigned chore must satisfy h[i] > x.

That means the valid integers are exactly:

L, L + 1, ..., R - 1

The count is:

R - L

If L == R, the answer naturally becomes 0.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(10^9 · n)	O(1)	Too slow
Optimal	O(n log n)	O(1) extra space apart from sorting	Accepted

Algorithm Walkthrough

1. Read the values n, a, and b, along with the list of chore difficulties.
2. Sort the difficulty array in nondecreasing order.

After sorting, the first b chores are the only possible chores Vasya can receive if he must take exactly b smallest acceptable chores. 3. Let:

L = h[b - 1]
R = h[b]

L is the largest difficulty assigned to Vasya, and R is the smallest difficulty assigned to Petya. 4. Every valid integer x must satisfy:

L ≤ x < R

The left condition guarantees Vasya keeps all his chores. The right condition guarantees Petya keeps all his chores. 5. The number of integers in that interval is:

R - L

6. Print R - L.

Why it works

After sorting, the division point between the two brothers is fixed. Vasya must receive exactly the first b chores, and Petya must receive exactly the remaining a chores.

For every chore assigned to Vasya, we need h[i] ≤ x. The strongest restriction comes from Vasya's hardest chore, which has value L.

For every chore assigned to Petya, we need h[i] > x. The strongest restriction comes from Petya's easiest chore, which has value R.

So the valid integers are precisely those between these two boundaries. Any smaller value excludes one of Vasya's chores, and any larger value includes one of Petya's chores.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n, a, b = map(int, input().split())
    h = list(map(int, input().split()))

    h.sort()

    print(h[b] - h[b - 1])

solve()

The solution relies entirely on the sorted order.

After sorting, index b - 1 is the last chore assigned to Vasya, while index b is the first chore assigned to Petya. Python uses zero-based indexing, so these positions are correct even though the problem describes the chores conceptually using one-based ordering.

The answer is simply the number of integers from h[b - 1] through h[b] - 1, which equals h[b] - h[b - 1].

A common off-by-one mistake is trying to use h[a - 1] and h[a]. The boundary is determined by Vasya's count, not Petya's count, because the first b sorted elements belong to Vasya.

Another easy mistake is subtracting one manually:

h[b] - h[b - 1] - 1

That would incorrectly exclude the left endpoint, even though x = h[b - 1] is valid.

Worked Examples

Sample 1

Input:

5 2 3
6 2 3 100 1

Sorted array:

[1, 2, 3, 6, 100]

Step	Value
b	3
h[b - 1]	3
h[b]	6
Answer	6 - 3 = 3

The valid integers are:

3, 4, 5

This trace shows that every integer inside the boundary gap works, not only values already present in the array.

Sample 2

Suppose the input is:

7 3 4
1 2 3 3 3 4 5

Sorted array:

[1, 2, 3, 3, 3, 4, 5]

Step	Value
b	4
h[b - 1]	3
h[b]	3
Answer	3 - 3 = 0

There is no integer x satisfying:

3 ≤ x < 3

This example demonstrates the duplicate-boundary edge case.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting dominates the runtime
Space	O(1) extra space apart from sorting	Only a few variables are used

With n ≤ 2000, the runtime is tiny. Sorting 2000 integers is effectively instantaneous within the given limits.

Test Cases

# helper: run solution on input string, return output string
import sys, io

def solve():
    input = sys.stdin.readline

    n, a, b = map(int, input().split())
    h = list(map(int, input().split()))

    h.sort()

    print(h[b] - h[b - 1])

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    backup = sys.stdout
    sys.stdout = out

    solve()

    sys.stdout = backup
    return out.getvalue()

# provided sample
assert run(
    "5 2 3\n"
    "6 2 3 100 1\n"
) == "3\n", "sample 1"

# minimum size
assert run(
    "2 1 1\n"
    "1 2\n"
) == "1\n", "minimum size"

# all equal values
assert run(
    "4 2 2\n"
    "5 5 5 5\n"
) == "0\n", "all equal"

# large gap
assert run(
    "5 2 3\n"
    "1 2 10 20 30\n"
) == "8\n", "multiple valid x values"

# boundary duplicate
assert run(
    "7 3 4\n"
    "1 2 3 3 3 4 5\n"
) == "0\n", "duplicate boundary"

# off-by-one check
assert run(
    "5 4 1\n"
    "1 2 3 4 5\n"
) == "1\n", "single valid value"

Test input	Expected output	What it validates
2 1 1 / 1 2	1	Minimum valid input
4 2 2 / 5 5 5 5	0	All values equal
5 2 3 / 1 2 10 20 30	8	Large interval of valid integers
7 3 4 / 1 2 3 3 3 4 5	0	Duplicate boundary values
5 4 1 / 1 2 3 4 5	1	Off-by-one boundary correctness

Edge Cases

Consider the duplicate-boundary case:

7 3 4
1 2 3 3 3 4 5

After sorting:

[1, 2, 3, 3, 3, 4, 5]

We compute:

L = h[3] = 3
R = h[4] = 3

The algorithm returns:

3 - 3 = 0

This is correct because no integer can simultaneously satisfy:

x ≥ 3
x < 3

Now consider the large-gap case:

5 2 3
1 2 10 20 30

After sorting:

[1, 2, 10, 20, 30]

We get:

L = 2
R = 10

The algorithm returns:

10 - 2 = 8

The valid integers are:

2, 3, 4, 5, 6, 7, 8, 9

This confirms that the answer counts every integer in the interval, not only difficulties already present in the array.

Finally, examine the smallest possible input:

2 1 1
1 2

After sorting:

[1, 2]

We compute:

L = 1
R = 2

The algorithm returns 1, meaning only x = 1 works.

If x = 0, Vasya receives no chores. If x = 2, Petya receives no chores. The boundary logic still works perfectly at minimum size.