1620. Coordinate With Maximum Network Quality

MediumArrayEnumeration

Problem Description

In this problem, we are given the locations and quality factors of several network towers. Each tower is represented as an array with three elements - the x-coordinate, the y-coordinate, and the quality factor. We have to find a point on an X-Y plane that has the maximum sum of the signal qualities from all the reachable towers.

A tower is considered reachable if it is within a given radius. The quality of the signal from a tower decreases as the distance from the tower increases, and it is calculated using the formula q_i / (1 + d), where q_i is the tower's quality factor and d is the Euclidean distance from the tower to the point of interest. We round down the calculated quality to the nearest integer.

We need to find the coordinates (cx, cy) where the network quality is the highest. If there are multiple such coordinates with the same quality, we return the smallest lexicographically coordinate, one that has the smallest non-negative x-coordinate, and if those are the same, then the one with the smallest non-negative y-coordinate.

Intuition

To solve this problem, we perform exhaustive search – basically, we iterate over all the possible coordinates and calculate the network quality at each point by summing up the signal quality from each tower that is within the given radius.

We initiate a variable mx to keep track of the maximum network quality found so far, and a variable ans to store the best coordinates corresponding to mx. We compute the network quality at every point by iterating over a predefined range (coordinates from (0, 0) to (50, 50) in this case, assuming these bounds are set by the problem context, i.e., the given x's and y's are expected to lie within this range).

For each point (i, j), we iterate over all towers checking their reachability. If a tower is reachable (distance d is less than or equal to radius), we calculate its contribution to the signal quality using the provided formula, rounding down to the nearest integer, and add it to the current network quality t.

If t exceeds mx, we update mx with t and set ans to the current coordinates (i, j). After checking all possible coordinates, ans will contain the coordinates with the highest possible network quality.

The reason for checking every point is that we aim to find the optimal location with maximum signal strength and we cannot deduce the best location without comparing the signal strengths at all possible locations due to the Euclidean distance factor and varying quality factors of the towers.

Solution Approach

The solution provided uses a brute-force approach to find the best coordinate with the maximum network quality.

For this, we use two nested loops to iterate over all possible coordinate points within the given constraints. In this case, it iterates through (0, 0) to (50, 50) as the solution only examines points with integral coordinates within this range.
We initiate a variable mx to track the highest network quality observed and a list ans to store the corresponding coordinates of this quality.
Inside the inner loop, for every point (i, j), another nested loop iterates through the list of towers. For each tower, it calculates the Euclidean distance d from the current coordinate (i, j) using the formula sqrt((x - i)^2 + (y - j)^2).
If the calculated distance is less than or equal to the radius, the tower is reachable, and its quality contribution is computed using the floor division formula q / (1 + d). This is because signal quality is integer-valued after flooring.
The variable t represents the total signal quality at the point (i, j) and is calculated by summing up the contributions from all reachable towers.
If at any point (i, j), the total signal quality t is greater than mx, mx is updated to t, and ans is updated to [i, j]. If t is equal to the maximum found so far, it automatically retains the earliest (and therefore lexicographically smallest) coordinates due to how the loops are structured (starting from (0, 0)).
The function floor() is used to perform the floor division operation to ensure that the quality is an integer.
Finally, after all the coordinates are checked, the function returns ans, which points to the coordinate with the highest network quality.

This approach does not use any sophisticated algorithmic pattern but relies on a simple exhaustive search. It is feasible when the range of coordinates is small, as the computational complexity is O(n * m * t), where n and m are the ranges of x and y coordinates to be checked, and t is the number of towers.

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Example Walkthrough

Consider the following small example to illustrate the solution approach:

Suppose we have two towers and the radius is given to be 2 units. The towers have the following properties:

Tower 1 has coordinates (1, 2) and a quality factor of 5.
Tower 2 has coordinates (3, 4) and a quality factor of 10.

To simplify, let's consider our grid extends from (0, 0) to (5, 5).

We would iterate from (0, 0) to (5, 5) and calculate the total network quality at each point. For example, at point (0, 0):

Compute the distance from Tower 1: d1 = sqrt((1 - 0)^2 + (2 - 0)^2) = sqrt(1 + 4) = sqrt(5)
Since d1 is greater than the radius of 2, Tower 1 is unreachable and contributes 0 to the network quality.
Compute the distance from Tower 2: d2 = sqrt((3 - 0)^2 + (4 - 0)^2) = sqrt(9 + 16) = sqrt(25)
Since d2 is also greater than the radius of 2, Tower 2 is unreachable and contributes 0 to the network quality.
The total network quality at point (0, 0) is 0.

Continuing this process for each point, assume we reach coordinate (2, 3):

Compute the distance from Tower 1: d1 = sqrt((1 - 2)^2 + (2 - 3)^2) = sqrt(1 + 1) = sqrt(2)
Tower 1 is reachable as d1 <= 2, so the quality contribution from Tower 1 is floor(5 / (1 + sqrt(2))).
Compute the distance from Tower 2: d2 = sqrt((3 - 2)^2 + (4 - 3)^2) = sqrt(1 + 1) = sqrt(2)
Tower 2 is reachable as d2 <= 2, so the quality contribution from Tower 2 is floor(10 / (1 + sqrt(2))).
The total network quality at point (2, 3) is the sum of quality contributions from both towers.

We continue this for all points on the grid from (0, 0) to (5, 5), keeping track of the maximum network quality and the corresponding coordinates. After we check all points:

Let's say the highest network quality found was 12 at point (2, 3). This would be our mx.
Given the highest quality, ans would be [2, 3].

Finally, the coordinates (2, 3) with the highest network quality 12 would be returned.

The reason for choosing such a small grid is to explain the brute force approach clearly without diving deep into a complex calculation that would involve many iterations. This step-by-step progression through the solution approach provides insight into the methodology of searching for the best signal quality on a grid.

Solution Implementation

1from math import floor
2
3class Solution:
4    def bestCoordinate(self, towers: List[List[int]], coverage_radius: int) -> List[int]:
5        # Initialize max signal quality and answer coordinates.
6        max_quality = 0
7        best_coordinate = [0, 0]
8      
9        # Iterate through each possible point on the grid.
10        for i in range(51):
11            for j in range(51):
12                total_quality = 0  # Sum of signal qualities from all towers for this point.
13              
14                # Calculate the signal quality from each tower at the current point.
15                for x, y, q in towers:
16                    # Calculate euclidean distance from the tower to the current point.
17                    distance = ((x - i) ** 2 + (y - j) ** 2) ** 0.5
18                  
19                    # Add quality to the total if inside the coverage radius.
20                    if distance <= coverage_radius:
21                        total_quality += floor(q / (1 + distance))
22              
23                # If the total signal quality at this point is greater than the max,
24                # update max_quality and best_coordinate to this point.
25                if total_quality > max_quality:
26                    max_quality = total_quality
27                    best_coordinate = [i, j]
28      
29        # After checking all points, return the coordinates with the highest signal quality.
30        return best_coordinate
31

1class Solution {
2    public int[] bestCoordinate(int[][] towers, int radius) {
3        int maxSignal = 0; // to keep track of the highest signal quality
4        int[] bestCoordinates = new int[]{0, 0}; // to hold the best coordinates
5      
6        // Loop through each possible coordinate on the grid up to 50x50
7        for (int x = 0; x < 51; x++) {
8            for (int y = 0; y < 51; y++) {
9                int signalQuality = 0; // to calculate the total signal quality at the point (x, y)
10              
11                // Check each tower's contribution to the signal quality at the point (x, y)
12                for (int[] tower : towers) {
13                    // Calculate the distance between tower and point (x, y)
14                    double distance = Math.sqrt(Math.pow(x - tower[0], 2) + Math.pow(y - tower[1], 2));
15                  
16                    // Add the tower's signal contribution if it is within radius
17                    if (distance <= radius) {
18                        signalQuality += Math.floor(tower[2] / (1 + distance));
19                    }
20                }
21              
22                // Update the maximum signal quality and the best coordinates if the current point is better
23                if (maxSignal < signalQuality) {
24                    maxSignal = signalQuality;
25                    bestCoordinates = new int[]{x, y};
26                }
27            }
28        }
29      
30        // Return the coordinates with the best signal quality
31        return bestCoordinates;
32    }
33}
34

1#include <vector>
2#include <cmath>
3using namespace std;
4
5class Solution {
6public:
7    // Function to find the best coordinate with the maximum signal quality
8    vector<int> bestCoordinate(vector<vector<int>>& towers, int radius) {
9        int maxSignalQuality = 0; // To store the maximum signal quality
10        vector<int> bestCoord = {0, 0}; // To store the best coordinate
11
12        // Iterate through each possible x coordinate within the grid limit
13        for (int x = 0; x < 51; ++x) {
14            // Iterate through each possible y coordinate within the grid limit
15            for (int y = 0; y < 51; ++y) {
16                int currentSignal = 0; // Signal quality at the current coordinate
17
18                // Iterate through each tower to calculate its contribution
19                for (auto& tower : towers) {
20                    // Calculate Euclidean distance from the current coordinate to the tower
21                    double distance = sqrt((x - tower[0]) * (x - tower[0]) + (y - tower[1]) * (y - tower[1]));
22                  
23                    // If the distance is within the effective radius, add the tower's signal quality
24                    if (distance <= radius) {
25                        currentSignal += static_cast<int>(floor(tower[2] / (1 + distance)));
26                    }
27                }
28
29                // Check if the current signal quality is greater than the max found so far
30                if (maxSignalQuality < currentSignal) {
31                    maxSignalQuality = currentSignal; // Update max signal quality
32                    bestCoord = {x, y}; // Update the best coordinate
33                }
34            }
35        }
36        // Return the best coordinate after checking all possible points
37        return bestCoord;
38    }
39};
40

1type Tower = [number, number, number];
2
3// Function to calculate Euclidean distance between two points
4function calculateDistance(x1: number, y1: number, x2: number, y2: number): number {
5  return Math.sqrt(Math.pow((x1 - x2), 2) + Math.pow((y1 - y2), 2));
6}
7
8// Function to find the best coordinate with the maximum signal quality
9function bestCoordinate(towers: Tower[], radius: number): [number, number] {
10  let maxSignalQuality: number = 0; // To store the maximum signal quality
11  let bestCoord: [number, number] = [0, 0]; // To store the best coordinate
12
13  // Iterate through each possible x coordinate within the grid limit
14  for (let x = 0; x < 51; ++x) {
15    // Iterate through each possible y coordinate within the grid limit
16    for (let y = 0; y < 51; ++y) {
17      let currentSignal: number = 0; // Signal quality at the current coordinate
18
19      // Iterate through each tower to calculate its contribution
20      for (let tower of towers) {
21        // Calculate Euclidean distance from the current coordinate to the tower
22        let distance: number = calculateDistance(x, y, tower[0], tower[1]);
23
24        // If the distance is within the effective radius, add the tower's signal quality
25        if (distance <= radius) {
26          currentSignal += Math.floor(tower[2] / (1 + distance));
27        }
28      }
29
30      // Check if the current signal quality is greater than the max found so far
31      if (maxSignalQuality < currentSignal) {
32        maxSignalQuality = currentSignal; // Update max signal quality
33        bestCoord = [x, y]; // Update the best coordinate
34      }
35    }
36  }
37
38  // Return the best coordinate after checking all possible points
39  return bestCoord;
40}
41

Time and Space Complexity

Time Complexity

The given code runs three nested loops:

The first two loops iterate over a fixed range of 51 each, resulting in 51 * 51 iterations.
The innermost loop iterates over the list of towers. If n is the total number of towers, this loop will execute n times for each iteration of the first two loops.

Therefore, the time complexity can be determined by multiplying the number of iterations in each loop. This gives us 51 * 51 * n, which simplifies to O(n) since the 51 * 51 is a constant.

Thus, the overall time complexity is O(n).

Space Complexity

The space complexity of the given code is quite straightforward. There are a few integers (mx, i, j, t, x, y, q, d) and a list ans of size 2 being used to store the answer, which do not depend on the input size.

Therefore, since no additional space is used that scales with the input size, the space complexity is O(1), or constant space complexity.

Learn more about how to find time and space complexity quickly using problem constraints.

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Question 1 out of 10

What does the following code do?

1def f(arr1, arr2):
2  i, j = 0, 0
3  new_arr = []
4  while i < len(arr1) and j < len(arr2):
5      if arr1[i] < arr2[j]:
6          new_arr.append(arr1[i])
7          i += 1
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1public static List<Integer> f(int[] arr1, int[] arr2) {
2  int i = 0, j = 0;
3  List<Integer> newArr = new ArrayList<>();
4
5  while (i < arr1.length && j < arr2.length) {
6      if (arr1[i] < arr2[j]) {
7          newArr.add(arr1[i]);
8          i++;
9      } else {
10          newArr.add(arr2[j]);
11          j++;
12      }
13  }
14
15  while (i < arr1.length) {
16      newArr.add(arr1[i]);
17      i++;
18  }
19
20  while (j < arr2.length) {
21      newArr.add(arr2[j]);
22      j++;
23  }
24
25  return newArr;
26}
27

1function f(arr1, arr2) {
2  let i = 0, j = 0;
3  let newArr = [];
4  
5  while (i < arr1.length && j < arr2.length) {
6      if (arr1[i] < arr2[j]) {
7          newArr.push(arr1[i]);
8          i++;
9      } else {
10          newArr.push(arr2[j]);
11          j++;
12      }
13  }
14  
15  while (i < arr1.length) {
16      newArr.push(arr1[i]);
17      i++;
18  }
19  
20  while (j < arr2.length) {
21      newArr.push(arr2[j]);
22      j++;
23  }
24  
25  return newArr;
26}
27

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