LeetCode #1620 — MEDIUM

Coordinate With Maximum Network Quality

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array of network towers towers, where towers[i] = [x_i, y_i, q_i] denotes the i^th network tower with location (x_i, y_i) and quality factor q_i. All the coordinates are integral coordinates on the X-Y plane, and the distance between the two coordinates is the Euclidean distance.

You are also given an integer radius where a tower is reachable if the distance is less than or equal to radius. Outside that distance, the signal becomes garbled, and the tower is not reachable.

The signal quality of the i^th tower at a coordinate (x, y) is calculated with the formula ⌊q_i / (1 + d)⌋, where d is the distance between the tower and the coordinate. The network quality at a coordinate is the sum of the signal qualities from all the reachable towers.

Return the array [c_x, c_y] representing the integral coordinate (c_x, c_y) where the network quality is maximum. If there are multiple coordinates with the same network quality, return the lexicographically minimum non-negative coordinate.

Note:

A coordinate (x1, y1) is lexicographically smaller than (x2, y2) if either:
- x1 < x2, or
- x1 == x2 and y1 < y2.
⌊val⌋ is the greatest integer less than or equal to val (the floor function).

Example 1:

Input: towers = [[1,2,5],[2,1,7],[3,1,9]], radius = 2
Output: [2,1]
Explanation: At coordinate (2, 1) the total quality is 13.
- Quality of 7 from (2, 1) results in ⌊7 / (1 + sqrt(0)⌋ = ⌊7⌋ = 7
- Quality of 5 from (1, 2) results in ⌊5 / (1 + sqrt(2)⌋ = ⌊2.07⌋ = 2
- Quality of 9 from (3, 1) results in ⌊9 / (1 + sqrt(1)⌋ = ⌊4.5⌋ = 4
No other coordinate has a higher network quality.

Example 2:

Input: towers = [[23,11,21]], radius = 9
Output: [23,11]
Explanation: Since there is only one tower, the network quality is highest right at the tower's location.

Example 3:

Input: towers = [[1,2,13],[2,1,7],[0,1,9]], radius = 2
Output: [1,2]
Explanation: Coordinate (1, 2) has the highest network quality.

Constraints:

1 <= towers.length <= 50
towers[i].length == 3
0 <= x_i, y_i, q_i <= 50
1 <= radius <= 50

Step 01

Brute Force Baseline

Problem summary: You are given an array of network towers towers, where towers[i] = [xi, yi, qi] denotes the ith network tower with location (xi, yi) and quality factor qi. All the coordinates are integral coordinates on the X-Y plane, and the distance between the two coordinates is the Euclidean distance. You are also given an integer radius where a tower is reachable if the distance is less than or equal to radius. Outside that distance, the signal becomes garbled, and the tower is not reachable. The signal quality of the ith tower at a coordinate (x, y) is calculated with the formula ⌊qi / (1 + d)⌋, where d is the distance between the tower and the coordinate. The network quality at a coordinate is the sum of the signal qualities from all the reachable towers. Return the array [cx, cy] representing the integral coordinate (cx, cy) where the network quality is maximum. If there are multiple

Baseline thinking

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

Pattern signal: Array

Example 1

[[1,2,5],[2,1,7],[3,1,9]]
2

Example 2

[[23,11,21]]
9

Example 3

[[1,2,13],[2,1,7],[0,1,9]]
2

Step 02

Core Insight

What unlocks the optimal approach

The constraints are small enough to consider every possible coordinate and calculate its quality.

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

Upper-end input sizes

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 #1620: Coordinate With Maximum Network Quality
class Solution {
    public int[] bestCoordinate(int[][] towers, int radius) {
        int mx = 0;
        int[] ans = new int[] {0, 0};
        for (int i = 0; i < 51; ++i) {
            for (int j = 0; j < 51; ++j) {
                int t = 0;
                for (var e : towers) {
                    double d = Math.sqrt((i - e[0]) * (i - e[0]) + (j - e[1]) * (j - e[1]));
                    if (d <= radius) {
                        t += Math.floor(e[2] / (1 + d));
                    }
                }
                if (mx < t) {
                    mx = t;
                    ans = new int[] {i, j};
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1620: Coordinate With Maximum Network Quality
func bestCoordinate(towers [][]int, radius int) []int {
	ans := []int{0, 0}
	mx := 0
	for i := 0; i < 51; i++ {
		for j := 0; j < 51; j++ {
			t := 0
			for _, e := range towers {
				d := math.Sqrt(float64((i-e[0])*(i-e[0]) + (j-e[1])*(j-e[1])))
				if d <= float64(radius) {
					t += int(float64(e[2]) / (1 + d))
				}
			}
			if mx < t {
				mx = t
				ans = []int{i, j}
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #1620: Coordinate With Maximum Network Quality
class Solution:
    def bestCoordinate(self, towers: List[List[int]], radius: int) -> List[int]:
        mx = 0
        ans = [0, 0]
        for i in range(51):
            for j in range(51):
                t = 0
                for x, y, q in towers:
                    d = ((x - i) ** 2 + (y - j) ** 2) ** 0.5
                    if d <= radius:
                        t += floor(q / (1 + d))
                if t > mx:
                    mx = t
                    ans = [i, j]
        return ans

// Accepted solution for LeetCode #1620: Coordinate With Maximum Network Quality
struct Solution;

use std::cmp::Reverse;

impl Solution {
    fn best_coordinate(towers: Vec<Vec<i32>>, radius: i32) -> Vec<i32> {
        let mut max = (0, Reverse(0), Reverse(0));
        for xi in 0..=50 {
            for yi in 0..=50 {
                let mut sum = 0;
                for point in towers.iter() {
                    let x = point[0];
                    let y = point[1];
                    let q = point[2];
                    if (x - xi) * (x - xi) + (y - yi) * (y - yi) > radius * radius {
                        continue;
                    }
                    let d = (((x - xi) * (x - xi) + (y - yi) * (y - yi)) as f64).sqrt();
                    sum += (q as f64 / (1.0 + d)).floor() as i32;
                }
                if (sum, Reverse(xi), Reverse(yi)) > max {
                    max = (sum, Reverse(xi), Reverse(yi));
                }
            }
        }
        vec![(max.1).0, (max.2).0]
    }
}

#[test]
fn test() {
    let towers = vec_vec_i32![[1, 2, 5], [2, 1, 7], [3, 1, 9]];
    let radius = 2;
    let res = vec![2, 1];
    assert_eq!(Solution::best_coordinate(towers, radius), res);
    let towers = vec_vec_i32![[23, 11, 21]];
    let radius = 9;
    let res = vec![23, 11];
    assert_eq!(Solution::best_coordinate(towers, radius), res);
    let towers = vec_vec_i32![[1, 2, 13], [2, 1, 7], [0, 1, 9]];
    let radius = 2;
    let res = vec![1, 2];
    assert_eq!(Solution::best_coordinate(towers, radius), res);
    let towers = vec_vec_i32![[2, 1, 9], [0, 1, 9]];
    let radius = 2;
    let res = vec![0, 1];
    assert_eq!(Solution::best_coordinate(towers, radius), res);
}

// Accepted solution for LeetCode #1620: Coordinate With Maximum Network Quality
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1620: Coordinate With Maximum Network Quality
// class Solution {
//     public int[] bestCoordinate(int[][] towers, int radius) {
//         int mx = 0;
//         int[] ans = new int[] {0, 0};
//         for (int i = 0; i < 51; ++i) {
//             for (int j = 0; j < 51; ++j) {
//                 int t = 0;
//                 for (var e : towers) {
//                     double d = Math.sqrt((i - e[0]) * (i - e[0]) + (j - e[1]) * (j - e[1]));
//                     if (d <= radius) {
//                         t += Math.floor(e[2] / (1 + d));
//                     }
//                 }
//                 if (mx < t) {
//                     mx = t;
//                     ans = new int[] {i, j};
//                 }
//             }
//         }
//         return ans;
//     }
// }

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.