Problème
The composite pattern treats single objects and groups of objects the same way. A folder's size is its own files plus the sizes of everything inside it, however deep.
There are n items numbered 0 to n − 1 forming a tree. parents[i] is the folder that contains item i, or -1 for the root (exactly one item). sizes[i] is the size of item i's own content.
Implement composite_folder_sizes(parents, sizes).
Input
parents: The parent of each item.
sizes: The own size of each item.
Output
Return, for every item, its total size: its own size plus the totals of all its children.
Contraintes
1 ≤ n ≤ 100,000
0 ≤ sizes[i] ≤ 10^9; parents describe a single tree
Exemples
Entrée :
sizes = [1, 2, 3, 4]
parents = [-1, 0, 0, 1]
Sortie attendue :
expected = [10, 6, 3, 4]Entrée :
std::vector<long long> sizes = {1, 2, 3, 4};
std::vector<long long> parents = {-1, 0, 0, 1};
Sortie attendue :
std::vector<long long> expected = {10, 6, 3, 4};Entrée :
long[] sizes = new long[] {1, 2, 3, 4};
long[] parents = new long[] {-1, 0, 0, 1};
Sortie attendue :
long[] expected = new long[] {10, 6, 3, 4};Explication
Item 3 (size 4) is inside item 1, so item 1 totals 2 + 4 = 6. The root holds items 1 and 2: 1 + 6 + 3 = 10. Result: [10, 6, 3, 4].
Entrée :
sizes = [5, 0, 7]
parents = [1, -1, 1]
Sortie attendue :
expected = [5, 12, 7]Entrée :
std::vector<long long> sizes = {5, 0, 7};
std::vector<long long> parents = {1, -1, 1};
Sortie attendue :
std::vector<long long> expected = {5, 12, 7};Entrée :
long[] sizes = new long[] {5, 0, 7};
long[] parents = new long[] {1, -1, 1};
Sortie attendue :
long[] expected = new long[] {5, 12, 7};Explication
The root is item 1 with no content of its own; it contains items 0 and 2: 0 + 5 + 7 = 12.
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
A folder's total is its own size plus its children's totals.
Indice 2
Process children before parents, for example in reverse depth order.
Indice 3
Avoid deep recursion on long chains; an explicit stack or ordering is safer.
Complexité attendue
O(n) time, O(n) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.