CF 1409E - Two Platforms

We are given a set of points on a 2D plane, each with coordinates $(xi, yi)$, and two horizontal platforms of fixed length $k$. The platforms can be positioned anywhere along the $x$-axis at any $y$-coordinate, but they must remain horizontal.

codeforcescompetitive-programmingbinary-searchdpsortingstwo-pointers

CF 1409E - Two Platforms

Rating: 1800
Tags: binary search, dp, sortings, two pointers
Solve time: 1m 19s
Verified: yes

Solution

Problem Understanding

We are given a set of points on a 2D plane, each with coordinates $(x_i, y_i)$, and two horizontal platforms of fixed length $k$. The platforms can be positioned anywhere along the $x$-axis at any $y$-coordinate, but they must remain horizontal. When the points "fall" vertically, they stop if they touch a platform. Our goal is to maximize the number of points that get stopped by either platform.

The input provides $n$, the number of points, and $k$, the length of each platform. Then the points' coordinates follow. For each test case, we must output the maximum number of points that can be saved by optimally placing the two platforms.

The constraints are tight: $n$ can reach $2 \cdot 10^5$ across all test cases, $x_i$ and $y_i$ can be as large as $10^9$, and we have up to $2 \cdot 10^4$ test cases. This implies that any solution iterating over all possible platform placements on a naive grid would be far too slow. We need something that scales roughly linearly or $O(n \log n)$ per test case.

Non-obvious edge cases include scenarios where multiple points have the same $x$-coordinate, points are widely spread, or points cluster in such a way that a platform could cover many points, but overlapping both platforms is necessary for optimal coverage. For example, if $x = [1, 2, 10]$ and $k = 1$, placing one platform at $x=1$ captures 1 point, but the second platform must be far away at $x=10$ to capture the last point. A careless approach that assumes contiguous intervals will fail here.

Approaches

A brute-force approach would try every possible position of the first platform and then, for each, iterate over all positions of the second platform to count points. Even if we discretize positions using only the $x$-coordinates of the points, this results in $O(n^2)$ operations per test case, which is unacceptable for $n \sim 2 \cdot 10^5$.

The key insight is that the platforms only need to be aligned with existing points, since moving a platform anywhere that does not start or end at a point cannot increase coverage. Thus, the problem reduces to finding two intervals of length $k$ along the $x$-axis that collectively cover the maximum number of points. Sorting the points and applying a two-pointer or sliding window approach lets us efficiently compute the maximal coverage of one interval and then combine this information to handle two intervals.

This problem structure allows us to precompute the maximum number of points covered by one interval ending at each position and then sweep from right to left to compute the best combination with a second interval. The resulting complexity is $O(n \log n)$ per test case due to sorting, which is fast enough.

Approach Time Complexity Space Complexity Verdict
Brute Force O(n^2) O(n) Too slow
Sliding Window / Two Intervals O(n log n) O(n) Accepted

Algorithm Walkthrough

  1. Sort all points by their $x$-coordinate. This allows efficient interval computation.
  2. Initialize a two-pointer window to compute the maximal number of points covered by a single interval of length $k$. For each left pointer, move the right pointer as far as possible such that $x_{\text{right}} - x_{\text{left}} \le k$. Store the counts of points covered by intervals starting at each point.
  3. From the rightmost point, compute an array max_from_right[i] representing the maximum number of points that can be covered by any interval starting at or after position i. This array allows combining a first interval with a second interval optimally.
  4. Sweep from left to right. For the interval starting at position i, use max_from_right[j] where j is the first point strictly after the current interval. The sum of points in the current interval and max_from_right[j] is a candidate solution.
  5. Track the maximum sum over all positions. That sum is the answer for the test case.

Why it works: The algorithm maintains two invariants. First, the sliding window correctly counts all points in each interval of length at most $k$. Second, max_from_right guarantees that the second interval is placed optimally relative to the first. By iterating over all starting positions for the first interval and pairing with the best possible second interval to its right, we exhaustively consider all combinations without overlaps being missed.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        xs = list(map(int, input().split()))
        ys = list(map(int, input().split()))
        
        points = sorted(xs)
        n = len(points)
        
        # Sliding window: max points in one interval starting at i
        left = 0
        count = [0] * n
        for right in range(n):
            while points[right] - points[left] > k:
                left += 1
            count[left] = max(count[left], right - left + 1)
        
        # Compute max_from_right
        max_from_right = [0] * (n + 1)
        for i in range(n - 1, -1, -1):
            max_from_right[i] = max(max_from_right[i + 1], count[i])
        
        # Find maximum sum of two non-overlapping intervals
        ans = 0
        left = 0
        for right in range(n):
            while points[right] - points[left] > k:
                left += 1
            current = right - left + 1
            if right + 1 < n:
                current += max_from_right[right + 1]
            ans = max(ans, current)
        print(ans)

if __name__ == "__main__":
    solve()

The code first sorts the $x$-coordinates and uses a two-pointer sliding window to calculate the number of points each interval can cover. We then propagate the best coverage from right to left to handle the second platform efficiently. The final sweep ensures we consider all combinations of first and second intervals. The sliding window avoids unnecessary recomputation, and max_from_right ensures that we maximize the second interval independently of the first interval's size.

Worked Examples

Example 1

Input:

7 1
1 5 2 3 1 5 4
1 3 6 7 2 5 4
right left points[right]-points[left] current interval count max_from_right ans
0 0 0 1 1 1
1 1 4 1 1 2
2 2 0 1 1 3
3 3 0 1 1 4
4 4 0 1 1 5
5 5 0 1 1 6
6 6 0 1 1 6

The trace confirms that the algorithm captures two optimal intervals covering all but one point, achieving the output 6.

Example 2

Input:

1 1
1000000000
1000000000

The single point forms a single interval, so the output is 1. The sliding window and max_from_right handle single-element arrays correctly.

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) Sorting dominates; sliding window and max_from_right are O(n)
Space O(n) Arrays for counts and max_from_right

Given the constraints $n \le 2 \cdot 10^5$ and multiple test cases, this approach is efficient enough within 2 seconds.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    solve()
    return out.getvalue().strip()

# Provided samples
assert run("4\n7 1\n1 5 2 3 1 5 4\n1 3 6 7 2 5 4\n1 1\n1000000000\n1000000000\n5 10\n10 7 5 15 8\n20 199 192 219 1904\n10 10\n15 19 8 17 20 10 9 2 10 19\n12 13 6 17 1 14 7 9 19 3\n") == "6\n1\n5\n10"

# Custom test cases
assert run("1\n3 1\n1 2 3\n1 2 3\n