How to deal out 2 different length arrays by the value? - arrays

Question

Think of a company, employees work in different capacities. Employees will be employed according to their capacities. There should be similar capacities for every day.
I want the array1's values ​​in the days array to be equally distributed.
For example: Thursday -> 2 item and 5 value, Friday -> 1 item and 5 value
Saturday -> 1 item and 6 value, Sunday -> 2 item and 6 value.
Let it be distributed with similar numbers.
I have 2 array like this:
let days = ["Thursday","Friday","Saturday","Sunday"]
var array1: [T] = [T(name: "a", value: 3),T(name: "b", value: 2),T(name: "c", value: 5),T(name: "d", value: 1),T(name: "e", value: 6),T(name: "f", value: 4)]
How can I deal out array1 objects to days array?
I tried this:
var arrayValueTotal = 0
for t in array1{
arrayValueTotal = arrayValueTotal + t.value
}
var upperLimit = Double(Double(arrayValueTotal)/Double(days.count)) // This will get value per day. 22/4 = 5.5
for i in (0...days.count-1){ //Handling the days
for j in array1{
if j.value >= Int(upperLimit){ //handling the array1's value
}else if j.value < Int(upperLimit){
}
}
}
Desired result like:
let combinedResult = [("Thursday": [T(name: "a", value: 3),
T(name: "b", value: 2)]),
("Friday": [T(name: "c", value: 5)]),
("Saturday": [T(name: "e", value: 6)]),
("Sunday": [T(name: "f", value: 5),
T(name: "d", value: 1)])]
Here is the Class T:
class T {
var name:String
var value:Int
init(name:String, value:Int) {
self.name = name
self.value = value
}
}
How can I get that? Thanks in advance.

Answer 1

Ok, if I understood correctly, let's start by separating array1 according to limit (that you calculated yourself):
var array1: [T] = [T(name: "a", value: 3),
T(name: "b", value: 2),
T(name: "c", value: 5),
T(name: "d", value: 1),
T(name: "e", value: 6),
T(name: "f", value: 4)]
let reducedArray = array1.reduce(into: [[T]]()) { result, current in
guard var last = result.last else { result.append([current]); return }
if last.reduce(0, { $0 + $1.value }) < limit {
last.append(current)
result[result.count - 1] = last
} else {
result.append([current])
}
}
Then reducedArray is:
[
[T(name:a, value: 3), T(name:b, value: 2)],
[T(name:c, value: 5)],
[T(name:d, value: 1), T(name:e, value: 6)],
[T(name:f, value: 4)]
]
Then
var final: [(String, [T])] = []
for i in 0..<min(days.count, reducedArray.count) { //If the count aren't equal, what to do with the left values?
final.append((days[i], reducedArray[i]))
}
print("Final: \(final)")
With final being:
[
("Thursday", [T(name:a, value: 3), T(name:b, value: 2)]),
("Friday", [T(name:c, value: 5)]),
("Saturday", [T(name:d, value: 1), T(name:e, value: 6)]),
("Sunday", [T(name:f, value: 4)])
]