The objective is to identify a specific coordinate along a linear axis that minimizes the cumulative distance to a predefined set of residential points. When mathematical models are constrained to selecting only from the existing data points, the solution converges toward the statistical median. Specifically, for an even number of observations, any point between the two central values yields an identical cost function; however, adhering to a minimum-value tie-breaker requires selecting the lower of the two median coordinates.

A deterministic approach involves evaluating every candidate coordinate from the input dataset. By calculating the aggregate displacement for each potential site and tracking the lowest observed metric, the optimal location can be isolated. Sorting the coordinates beforehand simplifies the tie-resolution mechanism, as iterating through ascending values ensures that the first encountered configuration matching the global minimum inherently satisfies the constraint for the smallest valid position.

Core Logic

1. Parse the integer count of households and thier respective linear coordinates.
2. Arrange the coordinate array in ascending order.
3. Initialize a tracker for the lowest aggregate distance and a placeholder for the target coordinate.
4. Execute a nested iteration where the outer loop selects a candidate site, and the inner loop sums the absolute deviations to all other sites.
5. Update the tracker when the computed aggregate distance is less than or equal to the current benchmark.
6. Output the final stored coordinate.

Java Solution

import java.util.Scanner;
import java.util.Arrays;

public class CampusPlacementCalculator {
    public static void main(String[] args) {
        Scanner inputReader = new Scanner(System.in);
        if (!inputReader.hasNextInt()) {
            return;
        }

        int totalHouseholds = inputReader.nextInt();
        int[] neighborhoodLocations = new int[totalHouseholds];

        for (int idx = 0; idx < totalHouseholds; idx++) {
            neighborhoodLocations[idx] = inputReader.nextInt();
        }

        Arrays.sort(neighborhoodLocations);

        long shortestAggregateDistance = Long.MAX_VALUE;
        int chosenSite = neighborhoodLocations[0];

        for (int i = 0; i < totalHouseholds; i++) {
            long currentTravelCost = 0;
            for (int j = 0; j < totalHouseholds; j++) {
                currentTravelCost += Math.abs(neighborhoodLocations[i] - neighborhoodLocations[j]);
            }

            if (currentTravelCost <= shortestAggregateDistance) {
                shortestAggregateDistance = currentTravelCost;
                chosenSite = neighborhoodLocations[i];
            }
        }

        System.out.println(chosenSite);
    }
}

Performance Characteristics

The algorithm operates with O(N²) time complexity due to the pairwise distance computations required for each candidate. Space complexity remains O(N) to store the input coordinates, which comfortab accommodates the maximum constraint of N=1000. For larger datasets, the computation can be reduced to O(N log N) by leveraging prefix-sum arrays or directly accessing the sorted median index, but the presented implementation prioritizes clarity and direct validation against the problem specifications.