CF 106017A - Permátomo tasters (Easy version)

The task asks us to look at every ordered pair of positions in an array. For a pair (i, j), we form a value by adding the two chosen elements. There are N² such values because choosing the first and second position are independent choices.

codeforcescompetitive-programming

CF 106017A - Perm\u00e1tomo tasters (Easy version)

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Solve time: 43s
Verified: yes

Solution

Problem Understanding

The task asks us to look at every ordered pair of positions in an array. For a pair (i, j), we form a value by adding the two chosen elements. There are N² such values because choosing the first and second position are independent choices. We need to find the K-th smallest value among all these pair sums.

The easy version allows N to be at most 500, with up to 20 test cases. This means a direct O(N²) approach is already around 250,000 operations per test case, which is acceptable. However, if we add extra work such as sorting all N² sums, the number of created values is also 250,000, and sorting them is still fine. The intended solution is usually to use the special structure of pair sums instead of materializing everything, because it teaches the technique used for larger variants too.

The values in the array can be as large as 10^9 in absolute value, so the answer can reach 2 * 10^9. A solution must use integer types that can safely hold these values. Python integers handle this automatically.

The first tricky case is when K refers to duplicate sums. For example:

1
3 4
1 1 5

The pair sums are:

2 2 6
2 2 6
6 6 10

The correct answer is 6 for the fourth smallest value. A solution that stores only unique sums would incorrectly return 10.

Another edge case appears with negative values:

1
2 2
-5 -1

The sums are -10, -6, -6, -2, so the answer is -6. A binary search that assumes the answer is positive or initializes its bounds incorrectly would fail here.

Approaches

The straightforward approach is to generate every possible couple. For every index i, we try every index j and append a[i] + a[j] into a list. After generating all N² sums, we sort the list and take the element at position K - 1.

This works because sorting places the complete multiset of sums in exactly the required order. The problem is that we create a large intermediate array and pay the sorting cost. The number of generated values is N², and sorting them costs O(N² log(N²)). For the easy limits this passes, but it does not scale well if the constraints grow.

The observation that improves the solution is that after sorting the original array, we can answer a different question: "How many pair sums are less than or equal to X?" If we can answer that quickly, we can binary search for the smallest X that contains at least K sums.

For a fixed value X, consider a sorted array. For each left element a[i], all valid partners form a suffix or prefix depending on the implementation. Since the array is sorted, we can use two pointers. Starting from the largest index on the right, if a[i] + a[j] is greater than X, we move j left. When it becomes valid, every index from 0 to j is also valid, adding j + 1 possible partners.

The brute force works because it explicitly sees every possible sum. The faster method works because sorted order lets us count many sums at once.

Approach Time Complexity Space Complexity Verdict
Brute Force O(N² log(N²)) O(N²) Accepted for easy version
Binary Search on Answer O(N log V + N log N) O(1) extra Accepted

Here V is the range of possible answers, about 2 * 10^9, so the binary search needs around 31 iterations.

Algorithm Walkthrough

  1. Sort the array. The sorted order lets us count how many pairs have a sum below a chosen value without checking every pair.
  2. Set the binary search range from the smallest possible pair sum to the largest possible pair sum. The smallest possible sum is created from the two smallest elements, and the largest possible sum comes from the two largest elements.
  3. For the current middle value mid, count how many ordered pairs (i, j) satisfy a[i] + a[j] <= mid.
  4. To count efficiently, keep a pointer j starting at the end of the sorted array. For every i, decrease j while the current pair is too large. Once a[i] + a[j] is valid, every index from 0 to j works with i, so add j + 1 to the count.
  5. If the count is at least K, the answer could be mid or smaller, so move the binary search right boundary down. Otherwise, not enough sums are small enough, so move the left boundary up.
  6. When the search ends, the left boundary is the smallest value that has at least K pair sums not exceeding it.

Why it works: the counting function is monotonic. If a value X has at least K sums not greater than it, every larger value also has at least K sums. If it has fewer than K, every smaller value also has fewer than K. Binary search finds the first value where the condition becomes true, which is exactly the K-th smallest sum.

Python Solution

import sys
input = sys.stdin.readline

def solve_case():
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    a.sort()

    def count_pairs(x):
        res = 0
        j = n - 1
        for i in range(n):
            while j >= 0 and a[i] + a[j] > x:
                j -= 1
            res += j + 1
        return res

    left = a[0] + a[0]
    right = a[-1] + a[-1]

    while left < right:
        mid = (left + right) // 2
        if count_pairs(mid) >= k:
            right = mid
        else:
            left = mid + 1

    return str(left)

def main():
    t = int(input())
    ans = []
    for _ in range(t):
        ans.append(solve_case())
    print("\n".join(ans))

if __name__ == "__main__":
    main()

The array is sorted first because the counting function depends on monotonic order. Without sorting, moving the pointer would not represent all possible partners.

The count_pairs function maintains a pointer j that only moves left during the whole scan. This is why the count is O(N) instead of O(N²). The value of j never resets for each new i.

The binary search uses left < right rather than searching over indices because we are looking for a value, not a position. The final value of left is the first sum threshold that contains at least K pairs.

The bounds are computed from the actual minimum and maximum possible sums. This handles negative numbers correctly and avoids assumptions about the input range.

Worked Examples

For the first sample:

2
3 4
1 3 5
2 2
-1 2

The first test case behaves as follows:

Step left right mid count(mid) Decision
Start 2 10
1 2 6 6 5 move right
End 2 6
Final 6 6 answer

The value 6 is the first threshold containing at least four pair sums.

For the second test case:

Step left right mid count(mid) Decision
Start -2 4
1 -2 1 1 3 move right
End -2 1 answer

The sorted array is [-1, 2], and the second smallest pair sum is 1.

Complexity Analysis

Measure Complexity Explanation
Time O(N log N + N log V) Sorting takes O(N log N), and each binary search step counts pairs in O(N)
Space O(1) extra Only the sorted array and a few counters are used

With N <= 500, the solution is easily within the limits. The same idea also scales to much larger arrays because it avoids building the full N² list.

Test Cases

import sys
import io

def solution(data):
    sys.stdin = io.StringIO(data)
    input = sys.stdin.readline

    def solve_case():
        n, k = map(int, input().split())
        a = list(map(int, input().split()))
        a.sort()

        def count_pairs(x):
            res = 0
            j = n - 1
            for i in range(n):
                while j >= 0 and a[i] + a[j] > x:
                    j -= 1
                res += j + 1
            return res

        l, r = a[0] + a[0], a[-1] + a[-1]
        while l < r:
            m = (l + r) // 2
            if count_pairs(m) >= k:
                r = m
            else:
                l = m + 1
        return str(l)

    t = int(input())
    out = []
    for _ in range(t):
        out.append(solve_case())
    return "\n".join(out)

assert solution("""2
3 4
1 3 5
2 2
-1 2
""") == "6\n1"

assert solution("""1
1 1
7
""") == "7"

assert solution("""1
4 8
2 2 2 2
""") == "4"

assert solution("""1
3 3
-5 -1 4
""") == "-1"

assert solution("""1
5 25
1 2 3 4 5
""") == "10"
Test input Expected output What it validates
Single element array 7 Minimum size handling
All values equal 4 Duplicate sums and repeated answers
Negative values -1 Correct binary search bounds
All pairs requested 10 Maximum K boundary

Edge Cases

For the duplicate case:

1
3 4
1 1 5

The sorted array remains [1, 1, 5]. The pair sums in sorted order are 2, 2, 2, 2, 6, 6, 6, 6, 6. During binary search, a threshold of 2 counts four pairs, so the answer becomes 2. The algorithm counts occurrences, not distinct values, so duplicates are handled naturally.

For the negative value case:

1
2 2
-5 -1

The search starts from -10 to -2. When checking -6, the counting step finds that the first element can pair with both elements, while the second element cannot exceed the threshold. The count reaches two, so the answer is -6. The negative range works because the bounds come directly from the array values.