Swift sorting nest String Array by Numbered Array Sort

Question

I have a Nested string array, below it's just an example a nested string array, This nested array cant be sorted alphabetically, because of that I created a Numbered array that corresponds to each nested string array.

To be clear the nested array cannot be sorted alphabetically so dont provide answers regarding that.

What Im trying to do, But don't understand how to achieve is to sort my int array from smallest to largest.

From there or at the same time short the nest array based on the Int arrays index changed

Nested String Array Example (Array Size Is Dynamic)

[["ddd"],["aaa"],["ggg"]]

Int Array Example (Array Size Is Dynamic)

[557, 147, 355]

After Sorting Both would look like this

[["aaa"],["ggg"], ["ddd"]]
[147, 355, 557]

Answer 1

You can zip both arrays together and sort it by the second element of the tuple. Then you just need to map the first element:

let a = [["ddd"],["aaa"],["ggg"]]
let b = [557, 147, 355]

let sorted = zip(a, b).sorted { $0.1 < $1.1 }

let sortedA = sorted.map(\.0)
sortedA // [["aaa"], ["ggg"], ["ddd"]]

let sortedB = sorted.map(\.1)
sortedB  // [147, 355, 557]

Answer 2

You can get an index from one array and sort using values from the second array:

let a = [["ddd"],["aaa"],["ggg"]]
let b = [557, 147, 355]

let sortedA = a.sorted { b[a.firstIndex(of: $0)!] < b[a.firstIndex(of: $1)!] }
// prints [["aaa"], ["ggg"], ["ddd"]]

let sortedB = b.sorted()
// prints [147, 355, 557]

---

This one-liner:

let sortedA = a.sorted { b[a.firstIndex(of: $0)!] < b[a.firstIndex(of: $1)!] }

is a shorter form of this:

let sortedA = a.sorted { item1, item2 in
    let item1Index = a.firstIndex(of: item1)!
    let item2Index = a.firstIndex(of: item2)!
    return b[item1Index] < b[item2Index]
}

Answer 3

@pawello2222's solution works but it has n squared performance, meaning it will be very slow for large data-sets.

Optimized sort algorithms tend to have n * log(n) performance, which is much, much better.

The following code will use the native sorted method in Swift, which should have good performance:

let a = [["ddd"],["aaa"],["ggg"]]
let b = [557, 147, 355]


let bIndexes = b

    //turn the array b into a new array of tuples 
    //containing elements and "offsets" (indexes)
    .enumerated() 

    .sorted { $0.element < $1.element }  //Sort the new array by elements
    .map { $0.offset }                   //Map the new array to just index.

print("bIndexes = \(bIndexes)")

//Loop through the array of indexes and use the indexes to fetch elements in sorted order.
bIndexes.forEach { print (b[$0], a[$0]) } 

This outputs:

bIndexes = [1, 2, 0]
147 ["aaa"]
355 ["ggg"]
557 ["ddd"]

Saving the results into sorted arrays is a simple matter of mapping:

let newA = bIndexes.map { a[$0] }
let newB = bIndexes.map { b[$0] }

Edit:

Or better yet, use LeoDabus' solution:

let a = [["ddd"],["aaa"],["ggg"]]
let b = [557, 147, 355]

let sortedIndexes = b.indices.sorted { b[$0] < b[$1] }

let sortedA = sortedIndexes.map { a[$0] }
let sortedB = sortedIndexes.map { b[$0] }

or if you only want the strings sorted based on the integer values:

let sortedA = b.indices.sorted { b[$0] < b[$1] }
.map { a[$0] }