Array to nested Dictionary in iOS? - ios

Question

There are a lot of questions with similar title but no one answers my question.
In my case I have an array:
["1", "2", "3", "4"]
And resulting dictionary should be:
["1": ["2": ["3": "4"]]]
How to convert it correctly? As I understand map, reduce and the similar functions doesn't have such functionality. And is it possible to do without of writing a recursive method?
Both Objective-C and Swift are applicable

Answer 1

A Swift solution, using reduce:
let a = ["1", "2", "3", "4"]
let d = a.dropLast(2).reversed()
.reduce([a[a.count - 2] : a[a.count - 1]]) { [$1 : $0] }
print(d) // ["1": ["2": ["3": "4"]]]
Here it is assumed that the array has at least 2 elements.
In our example, ["3", "4"] is the initial value for reduce,
and for the remaining elements in reverse order, a nested dictionary
is created in the closure with [$1 : $0].
The type of the result is [String: Any], which bridges to
NSDictionary if necessary.

Answer 2

//Your array
NSArray *a = #[#1,#2,#3,#4];
//Get the two last values (considering your array length must be at least 2 !
NSDictionary *last = #{a[a.count-2]:a[a.count-1]};
//In reverse order we assign last created dictionary to the current key, the array length must be superior than 3
for (NSInteger i = a.count - 3; i >= 0; i--)
last = #{a[i]:last};
NSLog(#"My dictionary %#", last);

Answer 3

Recursive option.
func toNestedDictionary(arr: [String]) -> [String: Any?]? {
if arr.isEmpty {
return nil
}
else if arr.count == 1 {
return Dictionary(dictionaryLiteral: (arr.first!, nil))
}
else if arr.count == 2 {
return Dictionary(dictionaryLiteral: (arr.first!, arr.last!))
}
else {
let head = arr.first!
let tail = arr.dropFirst()
return Dictionary(dictionaryLiteral: (head, toNestedDictionary(arr: Array(tail))))
}
}
print (toNestedDictionary(arr: ["1"]))
// Optional(["1": nil])
print (toNestedDictionary(arr: ["1", "2"]))
// Optional(["1": Optional("2")])
print (toNestedDictionary(arr: ["1", "2", "3"]))
// Optional(["1": Optional(["2": Optional("3")])])
print (toNestedDictionary(arr: ["1", "2", "3", "4"]))
// Optional(["1": Optional(["2": Optional(["3": Optional("4")])])])

Answer 4

Try below lines without recursion
NSArray *arr = #[#"1", #"2", #"3", #"4"];
NSMutableDictionary *dic = [[NSMutableDictionary alloc] init];
for (int i = arr.count-2; i>=0;i--) {
if (i == arr.count-2) {
[dic setObject:arr[i+1] forKey:arr[i]];
} else {
NSDictionary *tempDic = dic;
dic = #{arr[i]:tempDic};
}
}
NSLog(#"Solution%#",dic);

Answer 5

This problem can be solved using recursive programming easily. Here is the solution using Swift;
var ary = ["1", "2", "3", "4"]
func recursive(array:[String]) -> Any {
if ary.count > 1 {
let key = ary.removeFirst()
return [key : recursive(array: array)]
} else {
return array.last!
}
}
let reuslt = recursive(array: ary)
And result is: ["1": ["2": ["3": "4"]]]

Answer 6

Given a list
var list = ["1", "2", "3", "4"]
you can write
if let lastValue = list.popLast(), let lastKey = list.popLast() {
let dict = list.reversed().reduce([lastKey:lastValue]) { [$1:$0] }
print(dict)
}
Result
["1": ["2": ["3": "4"]]]
Please note that this code only works if the list contains at least 2 elements

Answer 7

Found a simplier solution:
NSMutableArray *tempArr = [array mutableCopy];
while (tempArray.count > 1) {
id value = tempArr.lastObject;
[tempArr removeLastObject];
NSString *key = tempArr.lastObject;
[tempArr removeLastObject];
[tempArr addObject:#{key: value}];
}
id result = tempArr.firstObject;