Sort complex array in swift

-5

I have a array like this:

 array = [[[2,2],[8,8]],[[4,4],[1,1]]]

I want to sort that array like this:

sortedArray = [[[1,1],[2,2]],[[4,4],[8,8]]]

Can anyone provide some kind of hint or solution for how to sort this array.

Thanks!

edited Nov 14 '18 at 12:30

fewlinesofcode

2,0671818

asked Nov 14 '18 at 12:27

Uday Babariya

547621

4

Are all elements in the sub-arrays equal? Are all powers of 2? Are they sorted? Do the sub-arrays have the same length?

– Cristik
Nov 14 '18 at 12:31

What if the array is [[[1,1],[9,9]],[4,4],[5,5]]] ?

– Rakesha Shastri
Nov 14 '18 at 12:33

Is it guaranteed that each nested array contains pair of equaled elements?

– Ahmad F
Nov 14 '18 at 12:33

1

You have like three-level nested arrays for what exactly? Could you enlighten a little more your problem?

– Guilherme Matuella
Nov 14 '18 at 12:35

You're going to have to move values out of these arrays into other array to... since 2 and 8 are within 1 array and 4 and 1 are in a different one so a sort based on the actual array is just going to throw back exactly what you already have as. As I suspect others will soon suggest, is there a reason for this amount of nesting, what are you actually trying to solve/store

– AngryDuck
Nov 14 '18 at 12:35

add a comment |

-5

I have a array like this:

 array = [[[2,2],[8,8]],[[4,4],[1,1]]]

I want to sort that array like this:

sortedArray = [[[1,1],[2,2]],[[4,4],[8,8]]]

Can anyone provide some kind of hint or solution for how to sort this array.

Thanks!

edited Nov 14 '18 at 12:30

fewlinesofcode

2,0671818

asked Nov 14 '18 at 12:27

Uday Babariya

547621

4

Are all elements in the sub-arrays equal? Are all powers of 2? Are they sorted? Do the sub-arrays have the same length?

– Cristik
Nov 14 '18 at 12:31

What if the array is [[[1,1],[9,9]],[4,4],[5,5]]] ?

– Rakesha Shastri
Nov 14 '18 at 12:33

Is it guaranteed that each nested array contains pair of equaled elements?

– Ahmad F
Nov 14 '18 at 12:33

1

You have like three-level nested arrays for what exactly? Could you enlighten a little more your problem?

– Guilherme Matuella
Nov 14 '18 at 12:35

You're going to have to move values out of these arrays into other array to... since 2 and 8 are within 1 array and 4 and 1 are in a different one so a sort based on the actual array is just going to throw back exactly what you already have as. As I suspect others will soon suggest, is there a reason for this amount of nesting, what are you actually trying to solve/store

– AngryDuck
Nov 14 '18 at 12:35

add a comment |

-5

I have a array like this:

 array = [[[2,2],[8,8]],[[4,4],[1,1]]]

I want to sort that array like this:

sortedArray = [[[1,1],[2,2]],[[4,4],[8,8]]]

Can anyone provide some kind of hint or solution for how to sort this array.

Thanks!

edited Nov 14 '18 at 12:30

fewlinesofcode

2,0671818

asked Nov 14 '18 at 12:27

Uday Babariya

547621

I have a array like this:

 array = [[[2,2],[8,8]],[[4,4],[1,1]]]

I want to sort that array like this:

sortedArray = [[[1,1],[2,2]],[[4,4],[8,8]]]

Can anyone provide some kind of hint or solution for how to sort this array.

Thanks!

swift

edited Nov 14 '18 at 12:30

fewlinesofcode

2,0671818

asked Nov 14 '18 at 12:27

Uday Babariya

547621

edited Nov 14 '18 at 12:30

fewlinesofcode

2,0671818

asked Nov 14 '18 at 12:27

Uday Babariya

547621

edited Nov 14 '18 at 12:30

fewlinesofcode

2,0671818

edited Nov 14 '18 at 12:30

fewlinesofcode

2,0671818

edited Nov 14 '18 at 12:30

fewlinesofcode

2,0671818

asked Nov 14 '18 at 12:27

Uday Babariya

547621

asked Nov 14 '18 at 12:27

Uday Babariya

547621

asked Nov 14 '18 at 12:27

Uday Babariya

547621

4

Are all elements in the sub-arrays equal? Are all powers of 2? Are they sorted? Do the sub-arrays have the same length?

– Cristik
Nov 14 '18 at 12:31

What if the array is [[[1,1],[9,9]],[4,4],[5,5]]] ?

– Rakesha Shastri
Nov 14 '18 at 12:33

Is it guaranteed that each nested array contains pair of equaled elements?

– Ahmad F
Nov 14 '18 at 12:33

1

You have like three-level nested arrays for what exactly? Could you enlighten a little more your problem?

– Guilherme Matuella
Nov 14 '18 at 12:35

You're going to have to move values out of these arrays into other array to... since 2 and 8 are within 1 array and 4 and 1 are in a different one so a sort based on the actual array is just going to throw back exactly what you already have as. As I suspect others will soon suggest, is there a reason for this amount of nesting, what are you actually trying to solve/store

– AngryDuck
Nov 14 '18 at 12:35

add a comment |

4

Are all elements in the sub-arrays equal? Are all powers of 2? Are they sorted? Do the sub-arrays have the same length?

– Cristik
Nov 14 '18 at 12:31

What if the array is [[[1,1],[9,9]],[4,4],[5,5]]] ?

– Rakesha Shastri
Nov 14 '18 at 12:33

Is it guaranteed that each nested array contains pair of equaled elements?

– Ahmad F
Nov 14 '18 at 12:33

1

You have like three-level nested arrays for what exactly? Could you enlighten a little more your problem?

– Guilherme Matuella
Nov 14 '18 at 12:35

You're going to have to move values out of these arrays into other array to... since 2 and 8 are within 1 array and 4 and 1 are in a different one so a sort based on the actual array is just going to throw back exactly what you already have as. As I suspect others will soon suggest, is there a reason for this amount of nesting, what are you actually trying to solve/store

– AngryDuck
Nov 14 '18 at 12:35

Are all elements in the sub-arrays equal? Are all powers of 2? Are they sorted? Do the sub-arrays have the same length?

– Cristik
Nov 14 '18 at 12:31

What if the array is [[[1,1],[9,9]],[4,4],[5,5]]] ?

– Rakesha Shastri
Nov 14 '18 at 12:33

Is it guaranteed that each nested array contains pair of equaled elements?

– Ahmad F
Nov 14 '18 at 12:33

You have like three-level nested arrays for what exactly? Could you enlighten a little more your problem?

– Guilherme Matuella
Nov 14 '18 at 12:35

You're going to have to move values out of these arrays into other array to... since 2 and 8 are within 1 array and 4 and 1 are in a different one so a sort based on the actual array is just going to throw back exactly what you already have as. As I suspect others will soon suggest, is there a reason for this amount of nesting, what are you actually trying to solve/store

– AngryDuck
Nov 14 '18 at 12:35

add a comment |

1 Answer
1

active

oldest

votes

Let's start with this array:

let array = [[[2,2],[8,8]],[[4,4],[1,1]]]

Flatten it, and sort it by the first element of each sub array:

let flat = array.flatMap { $0 }

    .sorted { $0.first ?? 0 < $1.first ?? 0 }

Now we'll need to group each element in an even index with the following element:

let start = flat.startIndex

let end = flat.endIndex

let even = stride(from: start, to: end, by: 2).map { flat[$0] }

let odd = stride(from: flat.index(start, offsetBy: 1), to: end, by: 2).map { flat[$0] }

let zipped = zip(even, odd)

let newArray = zipped.map { [$0.0, $0.1]}

And now you can use the result:

print(newArray) //[[[1, 1], [2, 2]], [[4, 4], [8, 8]]]

A more efficient way of creating pairs from elements at even and odd indices is to use this beautiful extension by matt:

extension Sequence {

    func clump(by clumpsize:Int) -> [[Element]] {

        let slices : [[Element]] = self.reduce(into:) {

            memo, cur in

            if memo.count == 0 {

                return memo.append([cur])

            }

            if memo.last!.count < clumpsize {

                memo.append(memo.removeLast() + [cur])

            } else {

                memo.append([cur])

            }

        }

        return slices

    }

}

And use it like so:

let array = [[[2,2],[8,8]],[[4,4],[1,1]]]

let flat = array.flatMap { $0 }

    .sorted { $0.first ?? 0 < $1.first ?? 0 }

let newArray = flat.clump(by: 2)

print(newArray) //[[[1, 1], [2, 2]], [[4, 4], [8, 8]]]

edited Nov 14 '18 at 14:21

answered Nov 14 '18 at 12:49

Carpsen90

7,65662760

perfect Answer..

– Uday Babariya
Nov 15 '18 at 11:30

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

Let's start with this array:

let array = [[[2,2],[8,8]],[[4,4],[1,1]]]

Flatten it, and sort it by the first element of each sub array:

let flat = array.flatMap { $0 }

    .sorted { $0.first ?? 0 < $1.first ?? 0 }

Now we'll need to group each element in an even index with the following element:

let start = flat.startIndex

let end = flat.endIndex

let even = stride(from: start, to: end, by: 2).map { flat[$0] }

let odd = stride(from: flat.index(start, offsetBy: 1), to: end, by: 2).map { flat[$0] }

let zipped = zip(even, odd)

let newArray = zipped.map { [$0.0, $0.1]}

And now you can use the result:

print(newArray) //[[[1, 1], [2, 2]], [[4, 4], [8, 8]]]

A more efficient way of creating pairs from elements at even and odd indices is to use this beautiful extension by matt:

extension Sequence {

    func clump(by clumpsize:Int) -> [[Element]] {

        let slices : [[Element]] = self.reduce(into:) {

            memo, cur in

            if memo.count == 0 {

                return memo.append([cur])

            }

            if memo.last!.count < clumpsize {

                memo.append(memo.removeLast() + [cur])

            } else {

                memo.append([cur])

            }

        }

        return slices

    }

}

And use it like so:

let array = [[[2,2],[8,8]],[[4,4],[1,1]]]

let flat = array.flatMap { $0 }

    .sorted { $0.first ?? 0 < $1.first ?? 0 }

let newArray = flat.clump(by: 2)

print(newArray) //[[[1, 1], [2, 2]], [[4, 4], [8, 8]]]

edited Nov 14 '18 at 14:21

answered Nov 14 '18 at 12:49

Carpsen90

7,65662760

perfect Answer..

– Uday Babariya
Nov 15 '18 at 11:30

add a comment |

Let's start with this array:

let array = [[[2,2],[8,8]],[[4,4],[1,1]]]

Flatten it, and sort it by the first element of each sub array:

let flat = array.flatMap { $0 }

    .sorted { $0.first ?? 0 < $1.first ?? 0 }

Now we'll need to group each element in an even index with the following element:

let start = flat.startIndex

let end = flat.endIndex

let even = stride(from: start, to: end, by: 2).map { flat[$0] }

let odd = stride(from: flat.index(start, offsetBy: 1), to: end, by: 2).map { flat[$0] }

let zipped = zip(even, odd)

let newArray = zipped.map { [$0.0, $0.1]}

And now you can use the result:

print(newArray) //[[[1, 1], [2, 2]], [[4, 4], [8, 8]]]

A more efficient way of creating pairs from elements at even and odd indices is to use this beautiful extension by matt:

extension Sequence {

    func clump(by clumpsize:Int) -> [[Element]] {

        let slices : [[Element]] = self.reduce(into:) {

            memo, cur in

            if memo.count == 0 {

                return memo.append([cur])

            }

            if memo.last!.count < clumpsize {

                memo.append(memo.removeLast() + [cur])

            } else {

                memo.append([cur])

            }

        }

        return slices

    }

}

And use it like so:

let array = [[[2,2],[8,8]],[[4,4],[1,1]]]

let flat = array.flatMap { $0 }

    .sorted { $0.first ?? 0 < $1.first ?? 0 }

let newArray = flat.clump(by: 2)

print(newArray) //[[[1, 1], [2, 2]], [[4, 4], [8, 8]]]

edited Nov 14 '18 at 14:21

answered Nov 14 '18 at 12:49

Carpsen90

7,65662760

perfect Answer..

– Uday Babariya
Nov 15 '18 at 11:30

add a comment |

Let's start with this array:

let array = [[[2,2],[8,8]],[[4,4],[1,1]]]

Flatten it, and sort it by the first element of each sub array:

let flat = array.flatMap { $0 }

    .sorted { $0.first ?? 0 < $1.first ?? 0 }

Now we'll need to group each element in an even index with the following element:

let start = flat.startIndex

let end = flat.endIndex

let even = stride(from: start, to: end, by: 2).map { flat[$0] }

let odd = stride(from: flat.index(start, offsetBy: 1), to: end, by: 2).map { flat[$0] }

let zipped = zip(even, odd)

let newArray = zipped.map { [$0.0, $0.1]}

And now you can use the result:

print(newArray) //[[[1, 1], [2, 2]], [[4, 4], [8, 8]]]

A more efficient way of creating pairs from elements at even and odd indices is to use this beautiful extension by matt:

extension Sequence {

    func clump(by clumpsize:Int) -> [[Element]] {

        let slices : [[Element]] = self.reduce(into:) {

            memo, cur in

            if memo.count == 0 {

                return memo.append([cur])

            }

            if memo.last!.count < clumpsize {

                memo.append(memo.removeLast() + [cur])

            } else {

                memo.append([cur])

            }

        }

        return slices

    }

}

And use it like so:

let array = [[[2,2],[8,8]],[[4,4],[1,1]]]

let flat = array.flatMap { $0 }

    .sorted { $0.first ?? 0 < $1.first ?? 0 }

let newArray = flat.clump(by: 2)

print(newArray) //[[[1, 1], [2, 2]], [[4, 4], [8, 8]]]

edited Nov 14 '18 at 14:21

answered Nov 14 '18 at 12:49

Carpsen90

7,65662760

Let's start with this array:

let array = [[[2,2],[8,8]],[[4,4],[1,1]]]

Flatten it, and sort it by the first element of each sub array:

let flat = array.flatMap { $0 }

    .sorted { $0.first ?? 0 < $1.first ?? 0 }

Now we'll need to group each element in an even index with the following element:

let start = flat.startIndex

let end = flat.endIndex

let even = stride(from: start, to: end, by: 2).map { flat[$0] }

let odd = stride(from: flat.index(start, offsetBy: 1), to: end, by: 2).map { flat[$0] }

let zipped = zip(even, odd)

let newArray = zipped.map { [$0.0, $0.1]}

And now you can use the result:

print(newArray) //[[[1, 1], [2, 2]], [[4, 4], [8, 8]]]

A more efficient way of creating pairs from elements at even and odd indices is to use this beautiful extension by matt:

extension Sequence {

    func clump(by clumpsize:Int) -> [[Element]] {

        let slices : [[Element]] = self.reduce(into:) {

            memo, cur in

            if memo.count == 0 {

                return memo.append([cur])

            }

            if memo.last!.count < clumpsize {

                memo.append(memo.removeLast() + [cur])

            } else {

                memo.append([cur])

            }

        }

        return slices

    }

}

And use it like so:

let array = [[[2,2],[8,8]],[[4,4],[1,1]]]

let flat = array.flatMap { $0 }

    .sorted { $0.first ?? 0 < $1.first ?? 0 }

let newArray = flat.clump(by: 2)

print(newArray) //[[[1, 1], [2, 2]], [[4, 4], [8, 8]]]

edited Nov 14 '18 at 14:21

answered Nov 14 '18 at 12:49

Carpsen90

7,65662760

edited Nov 14 '18 at 14:21

answered Nov 14 '18 at 12:49

Carpsen90

7,65662760

answered Nov 14 '18 at 12:49

Carpsen90

7,65662760

answered Nov 14 '18 at 12:49

Carpsen90

7,65662760

perfect Answer..

– Uday Babariya
Nov 15 '18 at 11:30

add a comment |

perfect Answer..

– Uday Babariya
Nov 15 '18 at 11:30

perfect Answer..

– Uday Babariya
Nov 15 '18 at 11:30

add a comment |

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