How best to find an element in nested lists? - functional-programming

Kotlin provides some usingful extension functions allow stream-like programming.
For example, if I look for an element in a list I can use find:
return list.find { n -> n>4 && n<6 }
But when I have a have nested lists this seems not practical for me. I have tu use forEach then -- luckyly I can return from an inner Lambda with Kotlin:
private fun findUsingForEach(data: List<List<Int>>, pred : (Int) -> Boolean) : Optional<Int> {
data.forEach { list ->
list.forEach { n ->
if( pred(n) ) return Optional.of(n)
}
}
return Optional.empty()
}
It seems fo me that forEach is not the right tool for that. Is there a more functional way to du this? filter comes to mind, but the nesting causes problems.
That follwing is the test I use for the function abouve:
#Test
open fun findTest() {
val data = listOf( listOf(1,2,3), listOf(3,4,5,6), listOf(), listOf(6,7,8) )
val e = findUsingForEach( data, { n -> n>4 && n < 6 } )
assertEquals(5, e.get())
}

You could flatten the list:
fun <T> Iterable<Iterable<T>>.flatten(): List<T> (source)
Returns a single list of all elements from all collections in the given collection.
val data = listOf(listOf(1, 2, 3), listOf(3, 4, 5, 6), listOf(), listOf(6, 7, 8))
data.flatten().find { n -> n > 4 && n < 6 }
This will return a single list with the elements of the sublists in order. Then you can use find as usual.
In your example,
{{1, 2, 3}, {3, 4, 5, 6}, {}, {6, 7, 8}}
becomes
{1, 2, 3, 3, 4, 5, 6, 6, 7, 8}
and the result of find on this list is 5.
However, this will create a new list. Take a look at the source of flatten:
/**
* Returns a single list of all elements from all collections in the given collection.
*/
public fun <T> Iterable<Iterable<T>>.flatten(): List<T> {
val result = ArrayList<T>()
for (element in this) {
result.addAll(element)
}
return result
}
If you want to save memory, create a Sequence from your list first:
data.asSequence()
and then perform your operations on this sequence:
data.asSequence().flatten().find { n -> n > 4 && n < 6 }
Side note: your predicate, n > 4 && n < 6, is simply equivalent to n == 5.

If you just want to reduce codes and you don't care much about efficiency, try this.
list.flatten().find { your pred here }
Or
list.flatMap { it }.find { your pred }
Or create a useful utility which doesn't create new lists (faster/lower memory taken):
inline fun <T> Iterable<Iterable<T>>.forEachEach(f: (T) -> Unit) =
forEach { it.forEach(f) }

Related

Invert Map<K, List<V>> to Map<V, K>

map = mapOf((2: [3,4,5]), (7: [22,33,44]))
need to convert this to
mapOf(3:2, 4:2, 5:2, 22:7, 33:7, 44:7)
tried using associate with forEach, not sure of the syntax
There might be some nicer syntax, but this should work well enough.
fun main() {
val map = mapOf(
2 to listOf(3, 4, 5),
7 to listOf(22, 33, 44)
)
val transformedMap = map.flatMap { entry ->
entry.value.map { it to entry.key }
}.toMap()
println(transformedMap)
}
Prints
{3=2, 4=2, 5=2, 22=7, 33=7, 44=7}
Note that the toMap function states
The returned map preserves the entry iteration order of the original collection. If any of two pairs would have the same key the last one gets added to the map.
So if you have the same value in two different lists, only the last one will be included in the map.
fun main() {
val map = mapOf(
2 to listOf(3, 4, 5),
7 to listOf(22, 33, 44),
8 to listOf(3)
)
val transformedMap = map.flatMap { entry ->
entry.value.map { it to entry.key }
}.toMap()
println(transformedMap)
}
Prints {3=8, 4=2, 5=2, 22=7, 33=7, 44=7}
Zymus' answer is correct, and is also what I would probably write.
However, if this is something that will be called often, you might want to extract it to a separate function that is more efficient.
fun <K, V> Map<K, Iterable<V>>.invert(): Map<V, K> {
val newMap = mutableMapOf<V, K>()
for ((key, iterable) in this) {
for (value in iterable) {
newMap[value] = key
}
}
return newMap
}
Usage:
fun main() {
val map = mapOf((2 to listOf(3, 4, 5)), (7 to listOf(22, 33, 44)))
val inverted = map.invert()
println(inverted)
}
Output:
{3=2, 4=2, 5=2, 22=7, 33=7, 44=7}
This is functionally equivalent to
map.flatMap { (key, values) -> values.map { it to key } }.toMap()
including the behaviour where if there are duplicate values in the original input, only the last one will be preserved as a new key. However, the flatMap version creates many temporary Lists (the number of original keys + 1) and many temporary Pairs (the number of original values), whereas this iterative version creates no extra objects.

flattening an array via the AST [duplicate]

I have a JavaScript array like:
[["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]]
How would I go about merging the separate inner arrays into one like:
["$6", "$12", "$25", ...]
ES2019
ES2019 introduced the Array.prototype.flat() method which you could use to flatten the arrays. It is compatible with most environments, although it is only available in Node.js starting with version 11, and not at all in Internet Explorer.
const arrays = [
["$6"],
["$12"],
["$25"],
["$25"],
["$18"],
["$22"],
["$10"]
];
const merge3 = arrays.flat(1); //The depth level specifying how deep a nested array structure should be flattened. Defaults to 1.
console.log(merge3);
Older browsers
For older browsers, you can use Array.prototype.concat to merge arrays:
var arrays = [
["$6"],
["$12"],
["$25"],
["$25"],
["$18"],
["$22"],
["$10"]
];
var merged = [].concat.apply([], arrays);
console.log(merged);
Using the apply method of concat will just take the second parameter as an array, so the last line is identical to this:
var merged = [].concat(["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]);
Here's a short function that uses some of the newer JavaScript array methods to flatten an n-dimensional array.
function flatten(arr) {
return arr.reduce(function (flat, toFlatten) {
return flat.concat(Array.isArray(toFlatten) ? flatten(toFlatten) : toFlatten);
}, []);
}
Usage:
flatten([[1, 2, 3], [4, 5]]); // [1, 2, 3, 4, 5]
flatten([[[1, [1.1]], 2, 3], [4, 5]]); // [1, 1.1, 2, 3, 4, 5]
There is a confusingly hidden method, which constructs a new array without mutating the original one:
var oldArray = [[1],[2,3],[4]];
var newArray = Array.prototype.concat.apply([], oldArray);
console.log(newArray); // [ 1, 2, 3, 4 ]
It can be best done by javascript reduce function.
var arrays = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]];
arrays = arrays.reduce(function(a, b){
return a.concat(b);
}, []);
Or, with ES2015:
arrays = arrays.reduce((a, b) => a.concat(b), []);
js-fiddle
Mozilla docs
There's a new native method called flat to do this exactly.
(As of late 2019, flat is now published in the ECMA 2019 standard, and core-js#3 (babel's library) includes it in their polyfill library)
const arr1 = [1, 2, [3, 4]];
arr1.flat();
// [1, 2, 3, 4]
const arr2 = [1, 2, [3, 4, [5, 6]]];
arr2.flat();
// [1, 2, 3, 4, [5, 6]]
// Flatten 2 levels deep
const arr3 = [2, 2, 5, [5, [5, [6]], 7]];
arr3.flat(2);
// [2, 2, 5, 5, 5, [6], 7];
// Flatten all levels
const arr4 = [2, 2, 5, [5, [5, [6]], 7]];
arr4.flat(Infinity);
// [2, 2, 5, 5, 5, 6, 7];
Most of the answers here don't work on huge (e.g. 200 000 elements) arrays, and even if they do, they're slow.
Here is the fastest solution, which works also on arrays with multiple levels of nesting:
const flatten = function(arr, result = []) {
for (let i = 0, length = arr.length; i < length; i++) {
const value = arr[i];
if (Array.isArray(value)) {
flatten(value, result);
} else {
result.push(value);
}
}
return result;
};
Examples
Huge arrays
flatten(Array(200000).fill([1]));
It handles huge arrays just fine. On my machine this code takes about 14 ms to execute.
Nested arrays
flatten(Array(2).fill(Array(2).fill(Array(2).fill([1]))));
It works with nested arrays. This code produces [1, 1, 1, 1, 1, 1, 1, 1].
Arrays with different levels of nesting
flatten([1, [1], [[1]]]);
It doesn't have any problems with flattening arrays like this one.
Update: it turned out that this solution doesn't work with large arrays. It you're looking for a better, faster solution, check out this answer.
function flatten(arr) {
return [].concat(...arr)
}
Is simply expands arr and passes it as arguments to concat(), which merges all the arrays into one. It's equivalent to [].concat.apply([], arr).
You can also try this for deep flattening:
function deepFlatten(arr) {
return flatten( // return shalowly flattened array
arr.map(x=> // with each x in array
Array.isArray(x) // is x an array?
? deepFlatten(x) // if yes, return deeply flattened x
: x // if no, return just x
)
)
}
See demo on JSBin.
References for ECMAScript 6 elements used in this answer:
Spread operator
Arrow functions
Side note: methods like find() and arrow functions are not supported by all browsers, but it doesn't mean that you can't use these features right now. Just use Babel — it transforms ES6 code into ES5.
You can use Underscore:
var x = [[1], [2], [3, 4]];
_.flatten(x); // => [1, 2, 3, 4]
Generic procedures mean we don't have to rewrite complexity each time we need to utilize a specific behaviour.
concatMap (or flatMap) is exactly what we need in this situation.
// concat :: ([a],[a]) -> [a]
const concat = (xs,ys) =>
xs.concat (ys)
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = f => xs =>
xs.map(f).reduce(concat, [])
// id :: a -> a
const id = x =>
x
// flatten :: [[a]] -> [a]
const flatten =
concatMap (id)
// your sample data
const data =
[["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]]
console.log (flatten (data))
foresight
And yes, you guessed it correctly, it only flattens one level, which is exactly how it should work
Imagine some data set like this
// Player :: (String, Number) -> Player
const Player = (name,number) =>
[ name, number ]
// team :: ( . Player) -> Team
const Team = (...players) =>
players
// Game :: (Team, Team) -> Game
const Game = (teamA, teamB) =>
[ teamA, teamB ]
// sample data
const teamA =
Team (Player ('bob', 5), Player ('alice', 6))
const teamB =
Team (Player ('ricky', 4), Player ('julian', 2))
const game =
Game (teamA, teamB)
console.log (game)
// [ [ [ 'bob', 5 ], [ 'alice', 6 ] ],
// [ [ 'ricky', 4 ], [ 'julian', 2 ] ] ]
Ok, now say we want to print a roster that shows all the players that will be participating in game …
const gamePlayers = game =>
flatten (game)
gamePlayers (game)
// => [ [ 'bob', 5 ], [ 'alice', 6 ], [ 'ricky', 4 ], [ 'julian', 2 ] ]
If our flatten procedure flattened nested arrays too, we'd end up with this garbage result …
const gamePlayers = game =>
badGenericFlatten(game)
gamePlayers (game)
// => [ 'bob', 5, 'alice', 6, 'ricky', 4, 'julian', 2 ]
rollin' deep, baby
That's not to say sometimes you don't want to flatten nested arrays, too – only that shouldn't be the default behaviour.
We can make a deepFlatten procedure with ease …
// concat :: ([a],[a]) -> [a]
const concat = (xs,ys) =>
xs.concat (ys)
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = f => xs =>
xs.map(f).reduce(concat, [])
// id :: a -> a
const id = x =>
x
// flatten :: [[a]] -> [a]
const flatten =
concatMap (id)
// deepFlatten :: [[a]] -> [a]
const deepFlatten =
concatMap (x =>
Array.isArray (x) ? deepFlatten (x) : x)
// your sample data
const data =
[0, [1, [2, [3, [4, 5], 6]]], [7, [8]], 9]
console.log (flatten (data))
// [ 0, 1, [ 2, [ 3, [ 4, 5 ], 6 ] ], 7, [ 8 ], 9 ]
console.log (deepFlatten (data))
// [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
There. Now you have a tool for each job – one for squashing one level of nesting, flatten, and one for obliterating all nesting deepFlatten.
Maybe you can call it obliterate or nuke if you don't like the name deepFlatten.
Don't iterate twice !
Of course the above implementations are clever and concise, but using a .map followed by a call to .reduce means we're actually doing more iterations than necessary
Using a trusty combinator I'm calling mapReduce helps keep the iterations to a minium; it takes a mapping function m :: a -> b, a reducing function r :: (b,a) ->b and returns a new reducing function - this combinator is at the heart of transducers; if you're interested, I've written other answers about them
// mapReduce = (a -> b, (b,a) -> b, (b,a) -> b)
const mapReduce = (m,r) =>
(acc,x) => r (acc, m (x))
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = f => xs =>
xs.reduce (mapReduce (f, concat), [])
// concat :: ([a],[a]) -> [a]
const concat = (xs,ys) =>
xs.concat (ys)
// id :: a -> a
const id = x =>
x
// flatten :: [[a]] -> [a]
const flatten =
concatMap (id)
// deepFlatten :: [[a]] -> [a]
const deepFlatten =
concatMap (x =>
Array.isArray (x) ? deepFlatten (x) : x)
// your sample data
const data =
[ [ [ 1, 2 ],
[ 3, 4 ] ],
[ [ 5, 6 ],
[ 7, 8 ] ] ]
console.log (flatten (data))
// [ [ 1. 2 ], [ 3, 4 ], [ 5, 6 ], [ 7, 8 ] ]
console.log (deepFlatten (data))
// [ 1, 2, 3, 4, 5, 6, 7, 8 ]
To flatten an array of single element arrays, you don't need to import a library, a simple loop is both the simplest and most efficient solution :
for (var i = 0; i < a.length; i++) {
a[i] = a[i][0];
}
To downvoters: please read the question, don't downvote because it doesn't suit your very different problem. This solution is both the fastest and simplest for the asked question.
Another ECMAScript 6 solution in functional style:
Declare a function:
const flatten = arr => arr.reduce(
(a, b) => a.concat(Array.isArray(b) ? flatten(b) : b), []
);
and use it:
flatten( [1, [2,3], [4,[5,[6]]]] ) // -> [1,2,3,4,5,6]
const flatten = arr => arr.reduce(
(a, b) => a.concat(Array.isArray(b) ? flatten(b) : b), []
);
console.log( flatten([1, [2,3], [4,[5],[6,[7,8,9],10],11],[12],13]) )
Consider also a native function Array.prototype.flat() (proposal for ES6) available in last releases of modern browsers. Thanks to #(Константин Ван) and #(Mark Amery) mentioned it in the comments.
The flat function has one parameter, specifying the expected depth of array nesting, which equals 1 by default.
[1, 2, [3, 4]].flat(); // -> [1, 2, 3, 4]
[1, 2, [3, 4, [5, 6]]].flat(); // -> [1, 2, 3, 4, [5, 6]]
[1, 2, [3, 4, [5, 6]]].flat(2); // -> [1, 2, 3, 4, 5, 6]
[1, 2, [3, 4, [5, 6]]].flat(Infinity); // -> [1, 2, 3, 4, 5, 6]
let arr = [1, 2, [3, 4]];
console.log( arr.flat() );
arr = [1, 2, [3, 4, [5, 6]]];
console.log( arr.flat() );
console.log( arr.flat(1) );
console.log( arr.flat(2) );
console.log( arr.flat(Infinity) );
You can also try the new Array.flat() method. It works in the following manner:
let arr = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]].flat()
console.log(arr);
The flat() method creates a new array with all sub-array elements concatenated into it recursively up to the 1 layer of depth (i.e. arrays inside arrays)
If you want to also flatten out 3 dimensional or even higher dimensional arrays you simply call the flat method multiple times. For example (3 dimensions):
let arr = [1,2,[3,4,[5,6]]].flat().flat().flat();
console.log(arr);
Be careful!
Array.flat() method is relatively new. Older browsers like ie might not have implemented the method. If you want you code to work on all browsers you might have to transpile your JS to an older version. Check for MDN web docs for current browser compatibility.
A solution for the more general case, when you may have some non-array elements in your array.
function flattenArrayOfArrays(a, r){
if(!r){ r = []}
for(var i=0; i<a.length; i++){
if(a[i].constructor == Array){
flattenArrayOfArrays(a[i], r);
}else{
r.push(a[i]);
}
}
return r;
}
What about using reduce(callback[, initialValue]) method of JavaScript 1.8
list.reduce((p,n) => p.concat(n),[]);
Would do the job.
const common = arr.reduce((a, b) => [...a, ...b], [])
You can use Array.flat() with Infinity for any depth of nested array.
var arr = [ [1,2,3,4], [1,2,[1,2,3]], [1,2,3,4,5,[1,2,3,4,[1,2,3,4]]], [[1,2,3,4], [1,2,[1,2,3]], [1,2,3,4,5,[1,2,3,4,[1,2,3,4]]]] ];
let flatten = arr.flat(Infinity)
console.log(flatten)
check here for browser compatibility
Please note: When Function.prototype.apply ([].concat.apply([], arrays)) or the spread operator ([].concat(...arrays)) is used in order to flatten an array, both can cause stack overflows for large arrays, because every argument of a function is stored on the stack.
Here is a stack-safe implementation in functional style that weighs up the most important requirements against one another:
reusability
readability
conciseness
performance
// small, reusable auxiliary functions:
const foldl = f => acc => xs => xs.reduce(uncurry(f), acc); // aka reduce
const uncurry = f => (a, b) => f(a) (b);
const concat = xs => y => xs.concat(y);
// the actual function to flatten an array - a self-explanatory one-line:
const flatten = xs => foldl(concat) ([]) (xs);
// arbitrary array sizes (until the heap blows up :D)
const xs = [[1,2,3],[4,5,6],[7,8,9]];
console.log(flatten(xs));
// Deriving a recursive solution for deeply nested arrays is trivially now
// yet more small, reusable auxiliary functions:
const map = f => xs => xs.map(apply(f));
const apply = f => a => f(a);
const isArray = Array.isArray;
// the derived recursive function:
const flattenr = xs => flatten(map(x => isArray(x) ? flattenr(x) : x) (xs));
const ys = [1,[2,[3,[4,[5],6,],7],8],9];
console.log(flattenr(ys));
As soon as you get used to small arrow functions in curried form, function composition and higher order functions, this code reads like prose. Programming then merely consists of putting together small building blocks that always work as expected, because they don't contain any side effects.
ES6 One Line Flatten
See lodash flatten, underscore flatten (shallow true)
function flatten(arr) {
return arr.reduce((acc, e) => acc.concat(e), []);
}
or
function flatten(arr) {
return [].concat.apply([], arr);
}
Tested with
test('already flatted', () => {
expect(flatten([1, 2, 3, 4, 5])).toEqual([1, 2, 3, 4, 5]);
});
test('flats first level', () => {
expect(flatten([1, [2, [3, [4]], 5]])).toEqual([1, 2, [3, [4]], 5]);
});
ES6 One Line Deep Flatten
See lodash flattenDeep, underscore flatten
function flattenDeep(arr) {
return arr.reduce((acc, e) => Array.isArray(e) ? acc.concat(flattenDeep(e)) : acc.concat(e), []);
}
Tested with
test('already flatted', () => {
expect(flattenDeep([1, 2, 3, 4, 5])).toEqual([1, 2, 3, 4, 5]);
});
test('flats', () => {
expect(flattenDeep([1, [2, [3, [4]], 5]])).toEqual([1, 2, 3, 4, 5]);
});
Using the spread operator:
const input = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]];
const output = [].concat(...input);
console.log(output); // --> ["$6", "$12", "$25", "$25", "$18", "$22", "$10"]
I recommend a space-efficient generator function:
function* flatten(arr) {
if (!Array.isArray(arr)) yield arr;
else for (let el of arr) yield* flatten(el);
}
// Example:
console.log(...flatten([1,[2,[3,[4]]]])); // 1 2 3 4
If desired, create an array of flattened values as follows:
let flattened = [...flatten([1,[2,[3,[4]]]])]; // [1, 2, 3, 4]
If you only have arrays with 1 string element:
[["$6"], ["$12"], ["$25"], ["$25"]].join(',').split(',');
will do the job. Bt that specifically matches your code example.
I have done it using recursion and closures
function flatten(arr) {
var temp = [];
function recursiveFlatten(arr) {
for(var i = 0; i < arr.length; i++) {
if(Array.isArray(arr[i])) {
recursiveFlatten(arr[i]);
} else {
temp.push(arr[i]);
}
}
}
recursiveFlatten(arr);
return temp;
}
A Haskellesque approach
function flatArray([x,...xs]){
return x ? [...Array.isArray(x) ? flatArray(x) : [x], ...flatArray(xs)] : [];
}
var na = [[1,2],[3,[4,5]],[6,7,[[[8],9]]],10];
fa = flatArray(na);
console.log(fa);
ES6 way:
const flatten = arr => arr.reduce((acc, next) => acc.concat(Array.isArray(next) ? flatten(next) : next), [])
const a = [1, [2, [3, [4, [5]]]]]
console.log(flatten(a))
ES5 way for flatten function with ES3 fallback for N-times nested arrays:
var flatten = (function() {
if (!!Array.prototype.reduce && !!Array.isArray) {
return function(array) {
return array.reduce(function(prev, next) {
return prev.concat(Array.isArray(next) ? flatten(next) : next);
}, []);
};
} else {
return function(array) {
var arr = [];
var i = 0;
var len = array.length;
var target;
for (; i < len; i++) {
target = array[i];
arr = arr.concat(
(Object.prototype.toString.call(target) === '[object Array]') ? flatten(target) : target
);
}
return arr;
};
}
}());
var a = [1, [2, [3, [4, [5]]]]];
console.log(flatten(a));
if you use lodash, you can just use its flatten method: https://lodash.com/docs/4.17.14#flatten
The nice thing about lodash is that it also has methods to flatten the arrays:
i) recursively: https://lodash.com/docs/4.17.14#flattenDeep
ii) upto n levels of nesting: https://lodash.com/docs/4.17.14#flattenDepth
For example
const _ = require("lodash");
const pancake = _.flatten(array)
I was goofing with ES6 Generators the other day and wrote this gist. Which contains...
function flatten(arrayOfArrays=[]){
function* flatgen() {
for( let item of arrayOfArrays ) {
if ( Array.isArray( item )) {
yield* flatten(item)
} else {
yield item
}
}
}
return [...flatgen()];
}
var flatArray = flatten([[1, [4]],[2],[3]]);
console.log(flatArray);
Basically I'm creating a generator that loops over the original input array, if it finds an array it uses the yield* operator in combination with recursion to continually flatten the internal arrays. If the item is not an array it just yields the single item. Then using the ES6 Spread operator (aka splat operator) I flatten out the generator into a new array instance.
I haven't tested the performance of this, but I figure it is a nice simple example of using generators and the yield* operator.
But again, I was just goofing so I'm sure there are more performant ways to do this.
just the best solution without lodash
let flatten = arr => [].concat.apply([], arr.map(item => Array.isArray(item) ? flatten(item) : item))
I would rather transform the whole array, as-is, to a string, but unlike other answers, would do that using JSON.stringify and not use the toString() method, which produce an unwanted result.
With that JSON.stringify output, all that's left is to remove all brackets, wrap the result with start & ending brackets yet again, and serve the result with JSON.parse which brings the string back to "life".
Can handle infinite nested arrays without any speed costs.
Can rightly handle Array items which are strings containing commas.
var arr = ["abc",[[[6]]],["3,4"],"2"];
var s = "[" + JSON.stringify(arr).replace(/\[|]/g,'') +"]";
var flattened = JSON.parse(s);
console.log(flattened)
Only for multidimensional Array of Strings/Numbers (not Objects)
Ways for making flatten array
using Es6 flat()
using Es6 reduce()
using recursion
using string manipulation
[1,[2,[3,[4,[5,[6,7],8],9],10]]] - [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
// using Es6 flat()
let arr = [1,[2,[3,[4,[5,[6,7],8],9],10]]]
console.log(arr.flat(Infinity))
// using Es6 reduce()
let flatIt = (array) => array.reduce(
(x, y) => x.concat(Array.isArray(y) ? flatIt(y) : y), []
)
console.log(flatIt(arr))
// using recursion
function myFlat(array) {
let flat = [].concat(...array);
return flat.some(Array.isArray) ? myFlat(flat) : flat;
}
console.log(myFlat(arr));
// using string manipulation
let strArr = arr.toString().split(',');
for(let i=0;i<strArr.length;i++)
strArr[i]=parseInt(strArr[i]);
console.log(strArr)
I think array.flat(Infinity) is a perfect solution. But flat function is a relatively new function and may not run in older versions of browsers. We can use recursive function for solving this.
const arr = ["A", ["B", [["B11", "B12", ["B131", "B132"]], "B2"]], "C", ["D", "E", "F", ["G", "H", "I"]]]
const flatArray = (arr) => {
const res = []
for (const item of arr) {
if (Array.isArray(item)) {
const subRes = flatArray(item)
res.push(...subRes)
} else {
res.push(item)
}
}
return res
}
console.log(flatArray(arr))

collapse list of int to list of ranges in kotlin

I have a list of ints that needs to be compressed to list of int ranges without loosing any information (there must be a way to reverse this operation).
Currently I have:
val ints = listOf(8, 9, 45, 48, 49, 60, 61, 61, 62, 63, 3, 4, 5, 4, 5, 6)
val out = ints
.map { it..it }
.fold(mutableListOf(ints[0]..(ints[0] - 1)),
{ acc, next ->
val prev = acc.last()
if (prev.last + 1 == next.first) {
acc[acc.lastIndex] = prev.first..next.last
} else {
acc.add(next)
}
acc
}).toList()
That correctly produces:
[8..9, 45..45, 48..49, 60..61, 61..63, 3..5, 4..6]
There are two aspects I dislike in my solution though,
it does not work for empty list because of fold's initial value
it's quite verbose for kotlin. I have a feeling that this can be resolved in bit nicer way.
So, the question is how to fix 1 and/or 2?
Thanks in advance!
Since you actually mutate the acc and return the same list of ranges at all iterations of fold, you may not really need the fold, that is, forEach is enough.
Then, mapping each number to it..it seems to be redundant here.
Taking the two notes above into account leads to the following, a bit simplified, version of your solution:
val result = mutableListOf<IntRange>()
ints.forEach {
val lastRange = result.lastOrNull()
if (lastRange?.endInclusive == it - 1)
result[result.lastIndex] = lastRange.first..it
else
result += it..it
}
UPD: with the addition of buildList to the Kotlin standard library, you can rewrite the above as:
val result = buildList {
ints.forEach {
val last = lastOrNull()
if (last?.endInclusive == it -1) {
set(lastIndex, last.start..it)
} else {
add(it..it)
}
}
}
My solution doesn't look much different, but I was able to fix your empty list issue:
val out = ints.fold(mutableListOf<IntRange>()) { acc, next ->
acc.apply {
if(isNotEmpty() && last().endInclusive.inc() == next) {
this[lastIndex] = this[lastIndex].start .. next
} else {
add(next..next)
}
}
}
It's also a bit less mapping, and using apply takes away some of the verbosity and having to refer to acc at the end.

How to achieve a pure function with dynamic programming in Kotlin?

I've been trying to transform some of my code to pure functions to learn how to use Kotlin in a functional way, with this simple snippet of code I can't think of any way to make my calculateFibonacci function a pure function.
I'm aware of a potentially recursive solution but what about a potential stack overflow, does Kotlin implement Tail Call Optimization?
Example:
val fibonacciValues = hashMapOf<Int, BigInteger>(0 to BigInteger.ONE, 1 to BigInteger.ONE);
// * TODO investigate how to do dynamic programming with a pure function ** //
private fun calculateFibonacci(n: Int): BigInteger? {
if (fibonacciValues.contains(n)) {
return fibonacciValues.get(n)
} else {
val f = calculateFibonacci(n - 2)!!.add(calculateFibonacci(n - 1))
fibonacciValues.put(n, f)
return f
}
}
For the whole snippet I uploaded this gist:
https://gist.github.com/moxi/e30f8e693bf044e8b6b80f8c05d4ac12
The whole thing is about breaking out of the imperative approach and thinking in terms of sequence manipulation.
In the case of the Fibonacci Sequence, it might be tricky because it's very tempting to think of it as a sequence of Ints but it gets much easier if you think of it as a sequence of pairs (from which you eventually derive a sequence of Ints)
So, you could create an infinite sequence of pairs where the next pair is defined as the second element of the previous pair and a sum of elements in a previous pair:
generateSequence(1 to 1) { it.second to it.first + it.second }
.map { it.first }
And yes, you can utilize the Tail Call Optimization by marking your method with the tailrec keyword - no worries about the stack overflow. You just apply it before the fun keyword:
fun fibonacciAt(n: Int) = {
tailrec fun fibonacciAcc(n: Int, a: Long, b: Long): Long {
return when (n == 0) {
true -> b
false -> fibonacciAcc(n - 1, a + b, a)
}
}
fibonacciAcc(n, 1, 0)
}
Here is more info about the Tail Recursion in Kotlin.
Homegrown:
fun fib(i: Int): Int {
tailrec fun go(k: Int, p: Int, c: Int): Int {
return if (k == 0) p
else go(k - 1, c, p + c)
}
return go(i, 0, 1)
}
generateSequence actually shows a Fibonacci implementation as example.
fun fibonacci(): Sequence<Int> {
// fibonacci terms
// 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, ...
return generateSequence(Pair(0, 1), { Pair(it.second, it.first + it.second) }).map { it.first }
}
println(fibonacci().take(10).toList()) // [0, 1, 1, 2, 3, 5, 8, 13, 21, 34]
does Kotlin implements Tail Call Optimization
Yes, there is tailrec keyword for that.

Finding the even and odd numbers in a linked list by recursion?

Happy new years for everyone! :)
I could do better things in the first day of the year but i'm trying to implement the linked lists and recursion together.
I just thought that how I can achieve to write a function that calculates the even numbers in a linked list with recursion.
void List:: findingEvens(Node* n, Node*& newHead){
if(n == NULL)
return;
else
if(n-> data % 2 != 0)
findingEvens(n-> next);
else{
if(!newHead){
newHead = n;
}
else{
Node* temp =head;
for(;temp->next;temp=temp->next){
temp = temp -> next;
}
temp-> next = n;
findingEvents(n-> next);
}
}
}
The problem is that in my h class I add the following as it should be
void findingEvens(Node* n);
However this makes me error which says that error: ‘Node’ has not been declared
Actually I have a Node struct after the definition of this function in h class.
Is the implementation of the recursive function wrong?
Any help will be appreciated, happy new year again :)
void List:: findingEvens(Node* n, Node*& newHead){
if(n == NULL)
return;
else
if(n-> data % 2 != 0)
findingEvens(n-> next, newHead);
else{
// Push even node onto newHead list
newNode = new Node;
newNode->data = n->data;
newNode->next = newHead;
newHead = newNode;
findingEvens(n-> next, newHead);
}
}
You need to pass newHead in the recursive calls.
You can't just assign n directly to newHead, because then the newHead list will have all the links from the original list. You need to make new nodes and copy the data.
The above code builds the result list in reverse order of the original list, e.g. if you start with 1, 2, 3, 5, 6, 8, 9, the result will be 8, 6, 2. You can reverse the list when it's done.

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