Calculate values of deep nested multi dimensional arrays in reverse - php

Question

I need to calculate reward for employees in my company.
I have this structure (for example):
And in PHP it represents the following array:
<?php
$structure = [
"A" => [
"B" => [
"E" => [
"M" => null
],
"F" => [
"N" => [
"T" => null
],
"O" => null
],
"G" => [
"P" => null,
"Q" => [
"U" => null,
"V" => [
"X" => null,
"Y" => [
"3" => [
"4" => [
"4" => null,
"6" => null,
"7" => [
"8" => null
],
]
]
],
"Z" => null
]
]
],
"H" => null
],
"C" => [
"I" => null,
"J" => [
"R" => null
],
"K" => [
"S" => [
"W" => [
"1" => null,
"2" => null,
]
]
]
],
"D" => [
"L" => null
]
]
];
I need to calculate the reward for each employee. The end subordinates have reward only from their own work. But the seniors who have other subordinates have reward from their own work + works their subordinates.
For example:
Person A has own reward 10.
Person D has own reward 20.
Person L has own reward 15.
In the final,
Person L has final reward 15 (is final).
Person D has final reward 20 + 15 = 35 (D + L).
Person A has final reward 10 + 35 (A + D).
The calculation must be carried out below, however, the network can be arbitrarily deep. Calculating I would like to split into several parts. (For performance reasons)
The spider, which revises the structure to the appropriate format.
Calculate reward for each node.
Send information about each node via email.
I do not know how to proceed across the structure. Or to reorganize the structure to undergo easier. Can you think of anything?
I am grateful for you. Thanks!
#Martin
// EDIT: raw database
| id | parent | name
---------------------
| 1 | null | Martin
| 2 | null | Peter
| 3 | 1 | John
| 4 | 3 | Jack
// EDIT: new data structure:
[
"A" => [
"points" => 20,
"childs" => [
"B" => [
"points" => 10,
"childs" => [
"C" => [
"points" => 50,
"childs" => null
]
]
],
"D" => [
"points" => 30,
"childs" => [
"4" => [
"points" => 40,
"childs" => null
]
]
]
]
]
]

Answer 1

This can be done by traversing through the structure recursively from the inside-out, and stores the rewards for each employee it can find in a flattened 2D array.
It is dependent on each leaf of the structure having a starting value, as it needs a base value to calculate back up the tree on.
Using RecursiveIteratorIterator with the CHILD_FIRST flag allows you to loop through the array 'backwards' which is what we want in this case as that is where the starting rewards are. As we go through the tree, we obtain the subordinates rewards, add it to the current employee, and continue .. so by the time we get back to the top of the structure, we have calculated all employees.
Storing the result in a flattened array is then much easier to use and manipulate further along your logic.
Assumed starting structure:
// Each leaf has a value (random for example sake)
$structure = [
"A" => [
"B" => [
"E" => [
"M" => 10
],
"F" => [
"N" => [
"T" => 15
],
"O" => 5
],
"G" => [
"P" => 40,
"Q" => [
"U" => 30,
"V" => [
"X" => 35,
"Y" => [
"3" => [
"4" => [
"5" => 5,
"6" => 10,
"7" => [
"8" => 20
],
]
]
],
"Z" => 30
]
]
],
"H" => 15
],
"C" => [
"I" => 25,
"J" => [
"R" => 25
],
"K" => [
"S" => [
"W" => [
"1" => 40,
"2" => 50,
]
]
]
],
"D" => [
"L" => 15
]
]
];
Calculation:
// Iterate through the structure from the outside-in (child/leaves first)
$data = new RecursiveArrayIterator($structure);
$dataIt = new RecursiveIteratorIterator($data, RecursiveIteratorIterator::CHILD_FIRST);
$rewards = [];
foreach ($dataIt as $value) {
$subKeys = [];
$rewards[$dataIt->key()] ?? $rewards[$dataIt->key()] = 0; // Suppress any undefined index errors
if (is_array($value)) {
$subKeys = array_keys($value);
// traverse through all branches to obtain the existing reward values for subordinates
array_walk_recursive($value, function($reward, $person) use (&$subKeys) {
$subKeys[] = $person;
});
$subKeys = array_unique($subKeys);
foreach ($subKeys as $employee) {
$rewards[$dataIt->key()] += $rewards[$employee];
}
} else {
$rewards[$dataIt->key()] += $value;
}
}
print_r($rewards);
Return/final array:
Array
(
[M] => 10
[E] => 10
[T] => 15
[N] => 15
[O] => 5
[F] => 35
[P] => 40
[U] => 30
[X] => 35
[4] => 60
[6] => 10
[8] => 20
[7] => 20
[3] => 90
[Y] => 180
[Z] => 30
[V] => 335
[Q] => 520
[G] => 745
[H] => 15
[B] => 1060
[I] => 25
[R] => 25
[J] => 25
[1] => 40
[2] => 50
[W] => 90
[S] => 180
[K] => 270
[C] => 435
[L] => 15
[D] => 15
[A] => 1935
)
To get the number of employees under each node, you can do that easily in one line using count:
$employeesPersonA = count($structure['A'], COUNT_RECURSIVE); // 33
$employeesPersonC = count($structure['A']['C'], COUNT_RECURSIVE); // 8
$employeesPersonK = count($structure['A']['C']['K'], COUNT_RECURSIVE); // 4
EDIT:
For your database structure, you cannot get a multi dimensional result set out of the database so your only option there is to go through the result set and build your structure in PHP from that.

Answer 2

Here's one example, simplified to consider only nodes, 'A','B','C',D', & 'L'...
DROP TABLE IF EXISTS my_table;
CREATE TABLE my_table
(user CHAR(1) NOT NULL PRIMARY KEY
,lft INT NOT NULL
,rgt INT NOT NULL
);
INSERT INTO my_table VALUES
('A',1,10),
('B',2,3),
('C',4,5),
('D',6,9),
('L',7,8);
SELECT y.user
, GROUP_CONCAT(x.user ORDER BY x.lft) nodes
FROM my_table x
JOIN my_table y
ON x.lft BETWEEN y.lft AND y.rgt
GROUP
BY y.user
ORDER
BY y.lft;
+------+-----------+
| user | nodes |
+------+-----------+
| A | A,B,C,D,L |
| B | B |
| C | C |
| D | D,L |
| L | L |
+------+-----------+