LeetCode #1515 — HARD

Best Position for a Service Centre

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

A delivery company wants to build a new service center in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new center in a position such that the sum of the euclidean distances to all customers is minimum.

Given an array positions where positions[i] = [x_i, y_i] is the position of the ith customer on the map, return the minimum sum of the euclidean distances to all customers.

In other words, you need to choose the position of the service center [x_centre, y_centre] such that the following formula is minimized:

Answers within 10^-5 of the actual value will be accepted.

Example 1:

Input: positions = [[0,1],[1,0],[1,2],[2,1]]
Output: 4.00000
Explanation: As shown, you can see that choosing [x_centre, y_centre] = [1, 1] will make the distance to each customer = 1, the sum of all distances is 4 which is the minimum possible we can achieve.

Example 2:

Input: positions = [[1,1],[3,3]]
Output: 2.82843
Explanation: The minimum possible sum of distances = sqrt(2) + sqrt(2) = 2.82843

Constraints:

1 <= positions.length <= 50
positions[i].length == 2
0 <= x_i, y_i <= 100

Step 01

Brute Force Baseline

Problem summary: A delivery company wants to build a new service center in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new center in a position such that the sum of the euclidean distances to all customers is minimum. Given an array positions where positions[i] = [xi, yi] is the position of the ith customer on the map, return the minimum sum of the euclidean distances to all customers. In other words, you need to choose the position of the service center [xcentre, ycentre] such that the following formula is minimized: Answers within 10-5 of the actual value will be accepted.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[[0,1],[1,0],[1,2],[2,1]]

Example 2

[[1,1],[3,3]]

Step 02

Core Insight

What unlocks the optimal approach

The problem can be reworded as, giving a set of points on a 2d-plane, return the geometric median.
Loop over each triplet of points (positions[i], positions[j], positions[k]) where i < j < k, get the centre of the circle which goes throw the 3 points, check if all other points lie in this circle.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1515: Best Position for a Service Centre
class Solution {
    public double getMinDistSum(int[][] positions) {
        int n = positions.length;
        double x = 0, y = 0;
        for (int[] p : positions) {
            x += p[0];
            y += p[1];
        }
        x /= n;
        y /= n;
        double decay = 0.999;
        double eps = 1e-6;
        double alpha = 0.5;
        while (true) {
            double gradX = 0, gradY = 0;
            double dist = 0;
            for (int[] p : positions) {
                double a = x - p[0], b = y - p[1];
                double c = Math.sqrt(a * a + b * b);
                gradX += a / (c + 1e-8);
                gradY += b / (c + 1e-8);
                dist += c;
            }
            double dx = gradX * alpha, dy = gradY * alpha;
            if (Math.abs(dx) <= eps && Math.abs(dy) <= eps) {
                return dist;
            }
            x -= dx;
            y -= dy;
            alpha *= decay;
        }
    }
}

// Accepted solution for LeetCode #1515: Best Position for a Service Centre
func getMinDistSum(positions [][]int) float64 {
	n := len(positions)
	var x, y float64
	for _, p := range positions {
		x += float64(p[0])
		y += float64(p[1])
	}
	x /= float64(n)
	y /= float64(n)
	const decay float64 = 0.999
	const eps float64 = 1e-6
	var alpha float64 = 0.5
	for {
		var gradX, gradY float64
		var dist float64
		for _, p := range positions {
			a := x - float64(p[0])
			b := y - float64(p[1])
			c := math.Sqrt(a*a + b*b)
			gradX += a / (c + 1e-8)
			gradY += b / (c + 1e-8)
			dist += c
		}
		dx := gradX * alpha
		dy := gradY * alpha
		if math.Abs(dx) <= eps && math.Abs(dy) <= eps {
			return dist
		}
		x -= dx
		y -= dy
		alpha *= decay
	}
}

# Accepted solution for LeetCode #1515: Best Position for a Service Centre
class Solution:
    def getMinDistSum(self, positions: List[List[int]]) -> float:
        n = len(positions)
        x = y = 0
        for x1, y1 in positions:
            x += x1
            y += y1
        x, y = x / n, y / n
        decay = 0.999
        eps = 1e-6
        alpha = 0.5
        while 1:
            grad_x = grad_y = 0
            dist = 0
            for x1, y1 in positions:
                a = x - x1
                b = y - y1
                c = sqrt(a * a + b * b)
                grad_x += a / (c + 1e-8)
                grad_y += b / (c + 1e-8)
                dist += c
            dx = grad_x * alpha
            dy = grad_y * alpha
            x -= dx
            y -= dy
            alpha *= decay
            if abs(dx) <= eps and abs(dy) <= eps:
                return dist

// Accepted solution for LeetCode #1515: Best Position for a Service Centre
struct Solution;
use rand::prelude::*;

impl Solution {
    fn get_min_dist_sum(positions: Vec<Vec<i32>>) -> f64 {
        let mut rng = thread_rng();
        let mut x = 50.0;
        let mut y = 50.0;
        let mut r = 50.0;
        let mut prev = Self::dist(&positions, x, y);
        for _ in 0..25 {
            for _ in 0..25 {
                let xx = x + rng.gen_range(-1.0, 1.0) * r;
                let yy = y + rng.gen_range(-1.0, 1.0) * r;
                let next = Self::dist(&positions, xx, yy);
                if next < prev {
                    prev = next;
                    x = xx;
                    y = yy;
                }
            }
            r *= 0.5;
        }
        prev
    }

    fn dist(positions: &[Vec<i32>], x: f64, y: f64) -> f64 {
        positions
            .iter()
            .map(|v| (((v[0] as f64 - x).powi(2) + (v[1] as f64 - y).powi(2)).sqrt()))
            .sum::<f64>()
    }
}

#[test]
fn test() {
    use assert_approx_eq::assert_approx_eq;
    let positions = vec_vec_i32![[0, 1], [1, 0], [1, 2], [2, 1]];
    let res = 4.0;
    assert_approx_eq!(Solution::get_min_dist_sum(positions), res, 1.0e-5);
    let positions = vec_vec_i32![[1, 1], [3, 3]];
    let res = 2.82843;
    assert_approx_eq!(Solution::get_min_dist_sum(positions), res, 1.0e-5);
    let positions = vec_vec_i32![[1, 1]];
    let res = 0.0;
    assert_approx_eq!(Solution::get_min_dist_sum(positions), res, 1.0e-5);
    let positions = vec_vec_i32![[1, 1], [0, 0], [2, 0]];
    let res = 2.73205;
    assert_approx_eq!(Solution::get_min_dist_sum(positions), res, 1.0e-5);
    let positions = vec_vec_i32![[0, 1], [3, 2], [4, 5], [7, 6], [8, 9], [11, 1], [2, 12]];
    let res = 32.94036;
    assert_approx_eq!(Solution::get_min_dist_sum(positions), res, 1.0e-5);
}

// Accepted solution for LeetCode #1515: Best Position for a Service Centre
function getMinDistSum(positions: number[][]): number {
    const n = positions.length;
    let [x, y] = [0, 0];
    for (const [px, py] of positions) {
        x += px;
        y += py;
    }
    x /= n;
    y /= n;
    const decay = 0.999;
    const eps = 1e-6;
    let alpha = 0.5;
    while (true) {
        let [gradX, gradY] = [0, 0];
        let dist = 0;
        for (const [px, py] of positions) {
            const a = x - px;
            const b = y - py;
            const c = Math.sqrt(a * a + b * b);
            gradX += a / (c + 1e-8);
            gradY += b / (c + 1e-8);
            dist += c;
        }
        const dx = gradX * alpha;
        const dy = gradY * alpha;
        if (Math.abs(dx) <= eps && Math.abs(dy) <= eps) {
            return dist;
        }
        x -= dx;
        y -= dy;
        alpha *= decay;
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.