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#!/bin/python3
import math
import os
// ... rest of solution available after purchaseExplanation
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To solve the problem of selecting k points from n distinct points on the x-axis so that the minimum distance between any two chosen points is as large as possible, you need to carefully analyze the problem and identify an approach that efficiently searches for the optimal spacing.
Step 1: Understand the problem
You have n distinct points given by their x-coordinates. You want to pick exactly k of these points. The goal is to maximize the minimum distance between any two points in the chosen set. This is a classic optimization problem often described as maximizing the minimum spacing or “aggressive cows” problem in algorithmic contexts.
Step 2: Sort the points
Since points are on the x-axis, their order matters. Begin by sorting the array of points in ascending order. This will make it easier to evaluate distances and systematically place points.
Step 3: Define the search space for the minimum distance
- The minimum distance you seek can range from 0 (or the smallest possible distance) to the maximum possible distance between the leftmost and rightmost points.
- The smallest possible minimum distance can be 0 if points are very close, and the largest possible minimum distance cannot exceed the difference between the maximum and minimum points.
- Thus, your search for the optimal minimum distance is bounded by this range.
Step 4: Use binary search on the minimum distance
The problem can be approached using binary search on the value of the minimum distance (let’s call it mid). For each candidate minimum distance mid, you check if it is possible to select k points from the sorted array such that every pair of chosen points is at least mid apart.
Step 5: Feasibility check (greedy approach)
For a given minimum distance mid:
- Start selecting points from the leftmost point (the smallest x-coordinate). This point is always chosen first.
- Then, move through the sorted points array and choose the next point only if its distance from the last chosen point is at least mid.
- Continue this process until you either select k points successfully (meaning mid is feasible) or run out of points before reaching k (meaning mid is not feasible).
This greedy selection works because if you can place k points with at least mid spacing in sorted order, any other selection won't be able to increase the minimum distance for that mid.
Step 6: Adjust binary search based on feasibility
- If for a candidate mid, you can successfully choose k points, it means mid is a feasible minimum distance and you try to increase the lower bound of your binary search to find a potentially larger minimum distance.
- If it’s not possible, reduce the upper bound to search for a smaller minimum distance.
- Continue narrowing down the search range until the bounds converge.
Step 7: Result interpretation
- After binary search completes, the lower bound will represent the maximum minimum distance achievable.
- This value guarantees that you can pick k points such that the smallest distance between any two of them is at least this value, and no larger minimum distance is feasible.
Step 8: Complexity considerations
- Sorting takes O(n log n).
- Each feasibility check takes O(n) because you iterate over the points once to greedily select k points.
- The binary search runs in O(log D), where D is the difference between max and min points.
- Overall complexity is O(n log n + n log D), which is efficient for typical input sizes.
Step 9: Edge cases
- When k = 2, the answer is simply the maximum distance between any two points (max - min).
- When k = n, the minimum distance is the smallest gap between any two consecutive points.
- Points that are very close together or very spread apart can influence feasibility, so careful handling of distances and comparisons is essential.
Summary:
This problem is best approached by sorting the points and performing a binary search on the minimum distance, combined with a greedy feasibility check to verify if that distance can be achieved with k points. This approach efficiently narrows down the answer to the maximum possible minimum distance.
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