Graphs
You are given a list of bombs. The range of a bomb is defined as the area where its effect can be felt. This area is in the shape of a circle with the center as the location of the bomb.
The bombs are represented by a 0-indexed 2D integer array
bombswherebombs[i] = [x[i], y[i], r[i]].x[i]andy[i]denote the X-coordinate and Y-coordinate of the location of the ith bomb, whereasr[i]denotes the radius of its range.You may choose to detonate a single bomb. When a bomb is detonated, it will detonate all bombs that lie in its range. These bombs will further detonate the bombs that lie in their ranges.
Given the list of
bombs, return the maximum number of bombs that can be detonated if you are allowed to detonate only one bomb.Constraints:
1 <= bombs.length <= 100bombs[i].length == 31 <= x[i], y[i], r[i] <= 10**5
Example 1
Input: bombs = [[2,1,3],[6,1,4]]
Output: 2
Explanation:
The above figure shows the positions and ranges of the 2 bombs.
If we detonate the left bomb, the right bomb will not be affected.
But if we detonate the right bomb, both bombs will be detonated.
So the maximum bombs that can be detonated is max(1, 2) = 2.
Example 2
Input: bombs = [[1,1,5],[10,10,5]]
Output: 1
Explanation:
Detonating either bomb will not detonate the other bomb, so the maximum number of bombs that can be detonated is 1.
Example 3
Input: bombs = [[1,2,3],[2,3,1],[3,4,2],[4,5,3],[5,6,4]]
Output: 5
Explanation:
The best bomb to detonate is bomb 0 because:
- Bomb 0 detonates bombs 1 and 2. The red circle denotes the range of bomb 0.
- Bomb 2 detonates bomb 3. The blue circle denotes the range of bomb 2.
- Bomb 3 detonates bomb 4. The green circle denotes the range of bomb 3.
Thus all 5 bombs are detonated.
It’s easy to think of the solution to count all points within each circle. If the radius is r, center is x0, y0, then r^2 >= (x - x0)^2 + (y - y0)^2 is a formula to check other points are within this circle or not. However, the problem introduced another bar of detonating one by one. Because of that, simple counting doesn’t work well. To represent the detonating situation, a graph data structure is used here. In the graph, each node (center) has adjacent nodes (neighboring centers) if the distance is less than or equals to the radius. Once the graph is constructed, do the breadth-first search (BFS) starting from each node (index) while counting up. The maximum count is the answer.
#include <queue>
#include <unordered_set>
#include <vector>
using namespace std;
class DetonateTheMaximumBombs {
public:
int maximumDetonation(vector<vector<int>>& bombs) {
auto isNeighbor = [](const vector<int> &a, const vector<int> &b) {
long long dx = a[0] - b[0], dy = a[1] - b[1];
long long r = a[2];
return r * r >= dx * dx + dy * dy;
};
int n = bombs.size();
// build a graph
vector<vector<int>> graph(n);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i == j) { continue; }
if (isNeighbor(bombs[i], bombs[j])) {
graph[i].push_back(j);
}
}
}
vector<int> counts(n, 1);
// bfs
for (int i = 0; i < n; ++i) {
queue<int> q({i});
unordered_set<int> seen({i});
while (!q.empty()) {
vector<int> neighbors = graph[q.front()];
q.pop();
for (int neigh : neighbors) {
if (seen.find(neigh) == seen.end()) {
counts[i]++;
seen.insert(neigh);
q.push(neigh);
}
}
}
}
return *max_element(counts.begin(), counts.end());
}
};
import java.util.*;
public class DetonateTheMaximumBombs {
private boolean isNeighbor(int[] a, int[] b) {
long dx = a[0] - b[0], dy = a[1] - b[1];
long r = a[2];
return r * r >= dx * dx + dy * dy;
}
public int maximumDetonation(int[][] bombs) {
int n = bombs.length;
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i < n; ++i) { graph.add(new ArrayList<>()); }
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i == j) { continue; }
if (isNeighbor(bombs[i], bombs[j])) {
graph.get(i).add(j);
}
}
}
int[] counts = new int[n];
for (int i = 0; i < n; ++i) { counts[i] = 1; }
for (int i = 0; i < n; ++i) {
Queue<Integer> q = new ArrayDeque<>();
q.add(i);
Set<Integer> seen = new HashSet<>();
seen.add(i);
while (!q.isEmpty()) {
List<Integer> neighbors = graph.get(q.poll());
for (int neigh : neighbors) {
if (!seen.contains(neigh)) {
counts[i]++;
seen.add(neigh);
q.add(neigh);
}
}
}
}
return Arrays.stream(counts).max().getAsInt();
}
}
/**
* @param {number[][]} bombs
* @return {number}
*/
var maximumDetonation = function(bombs) {
const isNeighbor = (a, b) => {
let dx = a[0] - b[0], dy = a[1] - b[1], r = a[2];
return r * r >= dx * dx + dy * dy;
};
const n = bombs.length;
const graph = [...Array(n)].map(_ => []);
for (let i = 0; i < n; i++) {
for (let j = 0; j < n; j++) {
if (i === j) { continue; }
if (isNeighbor(bombs[i], bombs[j])) {
graph[i].push(j);
}
}
}
const counts = new Array(n).fill(1);
for (let i = 0; i < n; i++) {
const q = [i];
const seen = new Set([i]);
while (q.length > 0) {
const neighbors = graph[q.shift()];
for (let neigh of neighbors) {
if (!seen.has(neigh)) {
counts[i] += 1;
seen.add(neigh);
q.push(neigh);
}
}
}
}
return Math.max(...counts);
};
class DetonateTheMaximumBombs:
def maximumDetonation(self, bombs: List[List[int]]) -> int:
def isNeighbor(a: List[int], b: List[int]) -> bool:
dx, dy = a[0] - b[0], a[1] - b[1]
return a[2] * a[2] >= dx * dx + dy * dy
n = len(bombs)
# build a graph
graph = [[] for _ in range(n)]
for i in range(n):
for j in range(n):
if i == j: continue
if isNeighbor(bombs[i], bombs[j]):
graph[i].append(j)
counts = [1 for _ in range(n)]
# bfs
for i in range(n):
queue, seen = [i], set({i})
while queue:
neighbors = graph[queue.pop(0)]
for neigh in neighbors:
if neigh not in seen:
counts[i] += 1
seen.add(neigh)
queue.append(neigh)
return max(counts)
require 'set'
# @param {Integer[][]} bombs
# @return {Integer}
def maximum_detonation(bombs)
is_neighbor = lambda do |a, b|
dx, dy, r = a[0] - b[0], a[1] - b[1], a[2]
r * r >= dx * dx + dy * dy
end
n = bombs.size
graph = Array.new(n){Array.new}
(0...n).each do |i|
(0...n).each do |j|
if i == j
next
end
if is_neighbor.call(bombs[i], bombs[j])
graph[i] << j
end
end
end
counts = Array.new(n).fill(1);
(0...n).each do |i|
q = [i]
seen = Set[i]
while !q.empty?
neighbors = graph[q.shift]
neighbors.each do |neigh|
if !seen.include?(neigh)
counts[i] += 1
seen.add(neigh)
q << neigh
end
end
end
end
counts.max
end
O(n * n)O(n)