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find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance.cpp 1.30 KB
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// Time: O(n^3)
// Space: O(n^2)
class Solution {
public:
int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
vector<vector<int>> dist(n, vector<int>(n, numeric_limits<int>::max()));
for (const auto& e : edges) {
dist[e[0]][e[1]] = dist[e[1]][e[0]] = e[2];
}
for (int i = 0; i < n; ++i) {
dist[i][i] = 0;
}
for (int k = 0; k < n; ++k) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (dist[i][k] != numeric_limits<int>::max() &&
dist[k][j] != numeric_limits<int>::max()) {
dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
}
}
}
}
int result = -1, min_num = numeric_limits<int>::max();
for (int i = n - 1; i >= 0; --i) {
int num = accumulate(dist[i].cbegin(), dist[i].cend(), 0,
[&distanceThreshold](const auto& a, const auto& b) {
return a + (b <= distanceThreshold);
});
if (num < min_num) {
min_num = num;
result = i;
}
}
return result;
}
};
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