The problem can be found at the following link: Question Link
- We solve this problem using Dijkstra's algorithm.
- We iterate over each city and find the count of cities that are reachable within the given threshold distance from that city using Dijkstra's algorithm.
- We keep track of the city with the smallest count of reachable cities and return it as the output.
- Time Complexity:
O((N + M)logN)
, where N is the number of cities and M is the number of edges. - Auxiliary Space Complexity:
O(N + M)
, where N is the number of cities and M is the number of edges.
class Solution {
public:
int dijkstra(int curr, int& n, int& distThreshold, vector<vector<pair<int, int>>>& graph) {
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
vector<int> vis(n, 0);
pq.push({0, curr});
int cnt = 0;
while (!pq.empty()) {
int dist = pq.top().first;
int node = pq.top().second;
pq.pop();
if (vis[node])
continue;
vis[node] = 1;
if (dist > distThreshold)
continue;
++cnt;
for (auto itr : graph[node]) {
int nodeDist = itr.first;
int nextNode = itr.second;
if (!vis[nextNode])
pq.push({dist + nodeDist, nextNode});
}
}
return cnt;
}
int findCity(int n, int m, vector<vector<int>>& edges, int distThreshold) {
vector<vector<pair<int, int>>> graph(n);
for (auto &itr : edges) {
graph[itr[0]].push_back({itr[2], itr[1]});
graph[itr[1]].push_back({itr[2], itr[0]});
}
int out = -1, nax = INT_MAX;
for (int i = 0; i < n; ++i) {
int cnt = dijkstra(i, n, distThreshold, graph);
if (cnt <= nax) {
nax = cnt;
out = i;
}
}
return out;
}
};
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