Hard
Parallel Courses II — C++
Full explanation · Time O((n * C(c, min(c, k))) * 2^n) · Space O(2^n)
// Time: O((n * C(c, min(c, k))) * 2^n)
// Space: O(2^n)
// concise dp solution
class Solution {
public:
int minNumberOfSemesters(int n, vector<vector<int>>& dependencies, int k) {
vector<int> reqs(n);
for (const auto& d : dependencies) {
reqs[d[1] - 1] |= 1 << (d[0] - 1);
}
vector<int> dp(1 << n, n);
dp[0] = 0;
for (int mask = 0; mask < dp.size(); ++mask) {
vector<int> candidates;
for (int v = 0; v < n; ++v) {
if ((mask & (1 << v)) == 0 && (mask & reqs[v]) == reqs[v]) {
candidates.emplace_back(v);
}
}
const auto r = min(int(candidates.size()), k);
combinations(candidates.size(), min(int(candidates.size()), k),
[&dp, &mask, &candidates](const vector<int>& idxs) {
auto new_mask = mask;
new_mask |= accumulate(cbegin(idxs), cend(idxs), 0,
[&candidates](const auto& a, const auto& b) {
return a | (1 << candidates[b]);
});
dp[new_mask] = min(dp[new_mask], dp[mask] + 1);
});
}
return dp.back();
}
private:
void combinations(int n, int k, const function<void (const vector<int>&)>& callback) {
static const auto& next_pos =
[](const auto& n, const auto& k, const auto& idxs) {
int i = k - 1;
for (; i >= 0; --i) {
if (idxs[i] != i + n - k) {
break;
}
}
return i;
};
vector<int> idxs(k);
iota(begin(idxs), end(idxs), 0);
callback(idxs);
for (int i; (i = next_pos(n, k, idxs)) >= 0;) {
++idxs[i];
for (int j = i + 1; j < k; ++j) {
idxs[j] = idxs[j - 1] + 1;
}
callback(idxs);
}
}
};
// Time: O((n * C(c, min(c, k))) * 2^n)
// Space: O(2^n)
// embedded combination dp solution
class Solution2 {
public:
int minNumberOfSemesters(int n, vector<vector<int>>& dependencies, int k) {
static const auto& choice_mask =
[](const auto& nums, const auto& idxs) {
return accumulate(cbegin(idxs), cend(idxs), 0,
[&nums](const auto& a, const auto& b) {
return a | (1 << nums[b]);
});
};
static const auto& next_pos =
[](const auto& n, const auto& r, const auto& idxs) {
int i = r - 1;
for (; i >= 0; --i) {
if (idxs[i] != i + n - r) {
break;
}
}
return i;
};
vector<int> reqs(n);
for (const auto& d : dependencies) {
reqs[d[1] - 1] |= 1 << (d[0] - 1);
}
vector<int> dp(1 << n, n);
dp[0] = 0;
for (int mask = 0; mask < dp.size(); ++mask) {
vector<int> candidates;
for (int v = 0; v < n; ++v) {
if ((mask & (1 << v)) == 0 && (mask & reqs[v]) == reqs[v]) {
candidates.emplace_back(v);
}
}
const auto r = min(int(candidates.size()), k);
vector<int> idxs(r);
iota(begin(idxs), end(idxs), 0);
const auto& new_mask = (mask | choice_mask(candidates, idxs));
dp[new_mask] = min(dp[new_mask], dp[mask] + 1);
for (int i; (i = next_pos(candidates.size(), r, idxs)) >= 0;) {
++idxs[i];
for (int j = i + 1; j < k; ++j) {
idxs[j] = idxs[j - 1] + 1;
}
const auto& new_mask = (mask | choice_mask(candidates, idxs));
dp[new_mask] = min(dp[new_mask], dp[mask] + 1);
}
}
return dp.back();
}
};
// Time: O(nlogn + e), e is the number of edges in graph
// Space: O(n + e)
// wrong greedy solution
// since the priority of courses are hard to decide especially for those courses with zero indegrees are of the same outdegrees and depths
// e.x.
// 9
// [[1,4],[1,5],[3,5],[3,6],[2,6],[2,7],[8,4],[8,5],[9,6],[9,7]]
// 3
class Solution_WA {
public:
int minNumberOfSemesters(int n, vector<vector<int>>& dependencies, int k) {
unordered_map<int, vector<int>> graph;
vector<int> degrees(n);
for (const auto &d: dependencies) {
graph[d[0] - 1].emplace_back(d[1] - 1);
++degrees[d[1] - 1];
}
vector<int> depths(n, -1);
for (int i = 0; i < n; ++i) {
dfs(graph, i, &depths);
}
priority_queue<pair<int, int>> max_heap;
for (int i = 0; i < n; ++i) {
if (!degrees[i]) {
max_heap.emplace(depths[i], i);
}
}
int result = 0;
while (!max_heap.empty()) {
vector<int> new_q;
for (int i = 0; !max_heap.empty() && i < k; ++i) {
const auto [depth, node] = max_heap.top(); max_heap.pop();
for (const auto& child : graph[node]) {
if (!--degrees[child]) {
new_q.emplace_back(child);
}
}
}
++result;
for (const auto& node : new_q) {
max_heap.emplace(depths[node], node);
}
}
return result;
}
private:
int dfs(const unordered_map<int, vector<int>> &graph,
int i, vector<int> *depths) {
if ((*depths)[i] == -1) {
int depth = 0;
if (graph.count(i)) {
for (const auto& child : graph.at(i)) {
depth = max(depth, dfs(graph, child, depths));
}
}
(*depths)[i] = depth + 1;
}
return (*depths)[i];
}
};