2101. Detonate the Maximum Bombs LeetCode Solution
In this guide, you will get 2101. Detonate the Maximum Bombs LeetCode Solution with the best time and space complexity. The solution to Detonate the Maximum Bombs problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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 bombs where bombs[i] = [xi, yi, ri]. xi and yi denote the X-coordinate and Y-coordinate of the location of the ith bomb, whereas ri 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.
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.
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.
2101. Detonate the Maximum Bombs LeetCode Solution in C++
class Solution {
public:
int maximumDetonation(vector<vector<int>>& bombs) {
const int n = bombs.size();
size_t ans = 0;
vector<vector<int>> graph(n);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i == j)
continue;
const long ri = bombs[i][2];
if (ri * ri >= squaredDist(bombs, i, j))
graph[i].push_back(j);
}
}
for (int i = 0; i < n; ++i) {
unordered_set<int> seen{i};
dfs(graph, i, seen);
ans = max(ans, seen.size());
}
return ans;
}
private:
void dfs(const vector<vector<int>>& graph, int u, unordered_set<int>& seen) {
for (const int v : graph[u]) {
if (seen.contains(v))
continue;
seen.insert(v);
dfs(graph, v, seen);
}
}
long squaredDist(const vector<vector<int>>& bombs, int i, int j) {
return static_cast<long>(bombs[i][0] - bombs[j][0]) *
(bombs[i][0] - bombs[j][0]) +
static_cast<long>(bombs[i][1] - bombs[j][1]) *
(bombs[i][1] - bombs[j][1]);
}
}; /* code provided by PROGIEZ */
2101. Detonate the Maximum Bombs LeetCode Solution in Java
class Solution {
public int maximumDetonation(int[][] bombs) {
final int n = bombs.length;
int ans = 0;
List<Integer>[] graph = new List[n];
for (int i = 0; i < n; ++i)
graph[i] = new ArrayList<>();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (i == j)
continue;
final long ri = bombs[i][2];
if (ri * ri >= squaredDist(bombs, i, j))
graph[i].add(j);
}
}
for (int i = 0; i < n; ++i) {
Set<Integer> seen = new HashSet<>(Arrays.asList(i));
dfs(graph, i, seen);
ans = Math.max(ans, seen.size());
}
return ans;
}
private void dfs(List<Integer>[] graph, int u, Set<Integer> seen) {
for (final int v : graph[u]) {
if (seen.contains(v))
continue;
seen.add(v);
dfs(graph, v, seen);
}
}
private long squaredDist(int[][] bombs, int i, int j) {
return (long) (bombs[i][0] - bombs[j][0]) * (bombs[i][0] - bombs[j][0]) +
(long) (bombs[i][1] - bombs[j][1]) * (bombs[i][1] - bombs[j][1]);
};
} // code provided by PROGIEZ
2101. Detonate the Maximum Bombs LeetCode Solution in Python
class Solution:
def maximumDetonation(self, bombs: list[list[int]]) -> int:
n = len(bombs)
ans = 0
graph = [[] for _ in range(n)]
for i, (xi, yi, ri) in enumerate(bombs):
for j, (xj, yj, rj) in enumerate(bombs):
if i == j:
continue
if ri**2 >= (xi - xj)**2 + (yi - yj)**2:
graph[i].append(j)
def dfs(u: int, seen: set[int]) -> None:
for v in graph[u]:
if v in seen:
continue
seen.add(v)
dfs(v, seen)
for i in range(n):
seen = set([i])
dfs(i, seen)
ans = max(ans, len(seen))
return ans # code by PROGIEZ
