1620. Coordinate With Maximum Network Quality
Description
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 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
, orx1 == x2
andy1 < y2
.
⌊val⌋
is the greatest integer less than or equal toval
(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 <= xi, yi, qi <= 50
1 <= radius <= 50
Solutions
Solution: Enumeration
- Time complexity: O(xyn)
- Space complexity: O(1)
JavaScript
/**
* @param {number[][]} towers
* @param {number} radius
* @return {number[]}
*/
const bestCoordinate = function (towers, radius) {
let minX = (minY = 50);
let maxX = (maxY = maxQ = 0);
let result = [0, 0];
towers.forEach(([x, y]) => {
minX = Math.min(x, minX);
minY = Math.min(y, minY);
maxX = Math.max(x, maxX);
maxY = Math.max(y, maxY);
});
for (let a = minX; a <= maxX; a++) {
for (let b = minY; b <= maxY; b++) {
let quality = 0;
for (const [x, y, q] of towers) {
const distance = Math.hypot((a - x), (b - y));
if (distance > radius) continue;
quality += Math.floor(q / (1 + distance));
}
if (quality <= maxQ) continue;
result = [a, b];
maxQ = quality;
}
}
return result;
};