Formatted question description: https://leetcode.ca/all/1203.html
1203. Sort Items by Groups Respecting Dependencies (Hard)
There are n items each belonging to zero or one of m groups where group[i] is the group that the i-th item belongs to and it's equal to -1 if the i-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it.
Return a sorted list of the items such that:
- The items that belong to the same group are next to each other in the sorted list.
- There are some relations between these items where
beforeItems[i]is a list containing all the items that should come before thei-th item in the sorted array (to the left of thei-th item).
Return any solution if there is more than one solution and return an empty list if there is no solution.
Example 1:
Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]] Output: [6,3,4,1,5,2,0,7]
Example 2:
Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]] Output: [] Explanation: This is the same as example 1 except that 4 needs to be before 6 in the sorted list.
Constraints:
1 <= m <= n <= 3*10^4group.length == beforeItems.length == n-1 <= group[i] <= m-10 <= beforeItems[i].length <= n-10 <= beforeItems[i][j] <= n-1i != beforeItems[i][j]beforeItems[i]does not contain duplicates elements.
Related Topics:
Depth-first Search, Graph, Topological Sort
Solution 1. Topological Sort
// OJ: https://leetcode.com/problems/sort-items-by-groups-respecting-dependencies/
// Time: O(M + N + E)
// Space: O(M + N + E)
class Solution {
public:
vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
unordered_map<int, vector<int>> itemGraph; // adjacency list of items. Edges between groups are considered in groupGraph so ignored here.
unordered_map<int, vector<int>> groupGraph; // adjacency list of groups.
vector<int> itemIndegree(n); // in-degree of each item
unordered_map<int, int> groupIndegree; // in-degree of each group
unordered_map<int, vector<int>> itemInGroup; // map from group id to items in the group
for (int i = 0; i < n; ++i) {
if (group[i] == -1) group[i] = i + m; // make items without group to be in group `i + m` so that we can treat all the items the same
itemInGroup[group[i]].push_back(i);
}
for (int i = 0; i < beforeItems.size(); ++i) {
for (int j = 0; j < beforeItems[i].size(); ++j) {
int from = beforeItems[i][j], to = i, fromGroup = group[from], toGroup = group[to];
if (fromGroup == toGroup) { // If the edge is in the same group, update itemGraph and itemIndegree
itemGraph[from].push_back(to);
itemIndegree[to]++;
} else { // If the edge is cross groups, update groupGraph and groupIndegree
groupGraph[fromGroup].push_back(toGroup);
groupIndegree[toGroup]++;
}
}
}
queue<int> groupQueue;
for (auto &p : itemInGroup) { // Get the initial set of groups without indegree
if (groupIndegree[p.first] == 0) groupQueue.push(p.first);
}
vector<int> ans;
while (groupQueue.size()) {
int gid = groupQueue.front();
groupQueue.pop();
queue<int> itemQueue; // do topological sort within the same group
int itemCnt = 0;
for (int u : itemInGroup[gid]) {
if (itemIndegree[u] == 0) itemQueue.push(u);
}
while (itemQueue.size()) {
int itemId = itemQueue.front();
itemQueue.pop();
ans.push_back(itemId);
++itemCnt;
for (int v : itemGraph[itemId]) {
if (--itemIndegree[v] == 0) itemQueue.push(v);
}
}
if (itemCnt != itemInGroup[gid].size()) return {}; // has circle within group, return empty list
for (int v : groupGraph[gid]) {
if (--groupIndegree[v] == 0) groupQueue.push(v);
}
}
if (ans.size() != n) return {}; // has circle between group, return empty list
return ans;
}
};
Solution 2. Topological Sort
// OJ: https://leetcode.com/problems/sort-items-by-groups-respecting-dependencies/
// Time: O(M + N + E)
// Space: O(M + N + E)
class Solution {
public:
vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
unordered_map<int, vector<int>> groupGraph, groupItems;
unordered_map<int, int> groupIndegree;
vector<int> groupOrder, ans;
for (int i = 0; i < n; ++i) {
int a = group[i] == -1 ? m + i : group[i];// For those items belonging to no group, let the groupId be `m + i`.
if (a < m) groupItems[group[i]].push_back(i);
if (groupGraph.count(a) == 0) groupGraph[a] = {};
for (int j : beforeItems[i]) {
int b = group[j] == -1 ? m + j : group[j];
if (a == b) continue; // skip intra dependency
groupGraph[b].push_back(a);
groupIndegree[a]++;
}
}
queue<int> q;
for (auto &[g, _] : groupGraph) {
if (groupIndegree[g] == 0) q.push(g);
}
while (q.size()) {
int u = q.front();
q.pop();
groupOrder.push_back(u);
for (int v : groupGraph[u]) {
if (--groupIndegree[v] == 0) q.push(v);
}
}
if (groupOrder.size() != groupGraph.size()) return {};
for (int g : groupOrder) {
if (g >= m) {
ans.push_back(g - m);
continue;
}
unordered_map<int, vector<int>> itemGraph;
unordered_map<int, int> itemIndegree;
for (int u : groupItems[g]) {
for (int v : beforeItems[u]) {
if (group[v] != g) continue;
itemGraph[v].push_back(u);
itemIndegree[u]++;
}
}
for (int u : groupItems[g]) {
if (itemIndegree[u] == 0) q.push(u);
}
int cnt = 0;
while (q.size()) {
int u = q.front();
q.pop();
ans.push_back(u);
++cnt;
for (int v : itemGraph[u]) {
if (--itemIndegree[v] == 0) q.push(v);
}
}
if (cnt != groupItems[g].size()) return {};
}
return ans;
}
};
Java
-
class Solution { public int[] sortItems(int n, int m, int[] group, List<List<Integer>> beforeItems) { for (int i = 0; i < n; i++) { if (group[i] == -1) { group[i] = m; m++; } } Map<Integer, Set<Integer>> groupOrderMap = new HashMap<Integer, Set<Integer>>(); Map<Integer, Set<Integer>> itemOrderMap = new HashMap<Integer, Set<Integer>>(); for (int i = 0; i < n; i++) { int currGroup = group[i]; List<Integer> beforeList = beforeItems.get(i); for (int before : beforeList) { int prevGroup = group[before]; Set<Integer> groupSet = groupOrderMap.getOrDefault(prevGroup, new HashSet<Integer>()); groupSet.add(currGroup); groupOrderMap.put(prevGroup, groupSet); Set<Integer> itemSet = itemOrderMap.getOrDefault(before, new HashSet<Integer>()); itemSet.add(i); itemOrderMap.put(before, itemSet); } } int[] groupIndegrees = new int[m]; for (int i = 0; i < m; i++) { Set<Integer> set = groupOrderMap.getOrDefault(i, new HashSet<Integer>()); for (int nextGroup : set) { if (nextGroup != i) groupIndegrees[nextGroup]++; } } Queue<Integer> groupQueue = new LinkedList<Integer>(); for (int i = 0; i < m; i++) { if (groupIndegrees[i] == 0) groupQueue.offer(i); } int[] groupsOrder = new int[m]; int groupIndex = 0; while (!groupQueue.isEmpty()) { int currGroup = groupQueue.poll(); groupsOrder[groupIndex] = currGroup; groupIndex++; Set<Integer> set = groupOrderMap.getOrDefault(currGroup, new HashSet<Integer>()); for (int nextGroup : set) { groupIndegrees[nextGroup]--; if (groupIndegrees[nextGroup] == 0) groupQueue.offer(nextGroup); } } if (groupIndex < m) return new int[0]; Map<Integer, Set<Integer>> groupItemsMap = new HashMap<Integer, Set<Integer>>(); for (int i = 0; i < n; i++) { int currGroup = group[i]; Set<Integer> set = groupItemsMap.getOrDefault(currGroup, new HashSet<Integer>()); set.add(i); groupItemsMap.put(currGroup, set); } int[] itemIndegrees = new int[n]; for (int i = 0; i < n; i++) { Set<Integer> set = itemOrderMap.getOrDefault(i, new HashSet<Integer>()); for (int nextItem : set) itemIndegrees[nextItem]++; } int[] itemsOrder = new int[n]; int itemIndex = 0; for (int i = 0; i < m; i++) { int currGroup = groupsOrder[i]; Set<Integer> itemSet = groupItemsMap.getOrDefault(currGroup, new HashSet<Integer>()); Queue<Integer> itemQueue = new LinkedList<Integer>(); for (int item : itemSet) { if (itemIndegrees[item] == 0) itemQueue.offer(item); } while (!itemQueue.isEmpty()) { int item = itemQueue.poll(); itemsOrder[itemIndex] = item; itemIndex++; Set<Integer> set = itemOrderMap.getOrDefault(item, new HashSet<Integer>()); for (int nextItem : set) { itemIndegrees[nextItem]--; if (itemIndegrees[nextItem] == 0 && group[nextItem] == currGroup) itemQueue.offer(nextItem); } } } if (itemIndex < n) return new int[0]; else return itemsOrder; } } -
Todo -
print("Todo!")