archanid's solution to List Ops on the TypeScript track

Instructions

Test suite

Solution

Note:

This exercise has changed since this solution was written.

Implement basic list operations.

In functional languages list operations like length, map, and reduce are very common. Implement a series of basic list operations, without using existing functions.

The precise number and names of the operations to be implemented will be track dependent to avoid conflicts with existing names, but the general operations you will implement include:

append (given two lists, add all items in the second list to the end of the first list);
concatenate (given a series of lists, combine all items in all lists into one flattened list);
filter (given a predicate and a list, return the list of all items for which predicate(item) is True);
length (given a list, return the total number of items within it);
map (given a function and a list, return the list of the results of applying function(item) on all items);
foldl (given a function, a list, and initial accumulator, fold (reduce) each item into the accumulator from the left using function(accumulator, item));
foldr (given a function, a list, and an initial accumulator, fold (reduce) each item into the accumulator from the right using function(item, accumulator));
reverse (given a list, return a list with all the original items, but in reversed order);

Setup

Go through the setup instructions for TypeScript to install the necessary dependencies:

https://exercism.io/tracks/typescript/installation

Requirements

Install assignment dependencies:

$ yarn install

Making the test suite pass

Execute the tests with:

$ yarn test

In the test suites all tests but the first have been skipped.

Once you get a test passing, you can enable the next one by changing xit to it.

Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

list-ops.test.ts

import List from './list-ops'

describe('append entries to a list and return the new list', () => {
    it('empty lists', () => {
        const list1 = new List()
        const list2 = new List()
        expect(list1.append(list2)).toEqual(new List())
    })

    xit('empty list to list', () => {
        const list1 = new List([1, 2, 3, 4])
        const list2 = new List<number>()
        expect(list1.append(list2)).toEqual(list1)
    })

    xit('non-empty lists', () => {
        const list1 = new List([1, 2])
        const list2 = new List([2, 3, 4, 5])
        expect(list1.append(list2).values).toEqual([1, 2, 2, 3, 4, 5])
    })
})

describe('concat lists and lists of lists into new list', () => {
    xit('empty list', () => {
        const list1 = new List()
        const list2 = new List([])
        expect(list1.concat(list2).values).toEqual([])
    })

    xit('list of lists', () => {
        const list1 = new List([1, 2])
        const list2 = new List([3])
        const list3 = new List([])
        const list4 = new List([4, 5, 6])
        const listOfLists = new List([list2, list3, list4])
        expect(list1.concat(listOfLists).values).toEqual([1, 2, 3, 4, 5, 6])
    })
})

describe('filter list returning only values that satisfy the filter function', () => {
    xit('empty list', () => {
        const list1 = new List([])
        expect(list1.filter((el: number) => el % 2 === 1).values).toEqual([])
    })

    xit('non empty list', () => {
        const list1 = new List([1, 2, 3, 5])
        expect(list1.filter((el: number) => el % 2 === 1).values).toEqual([1, 3, 5])
    })
})

describe('returns the length of a list', () => {
    xit('empty list', () => {
        const list1 = new List()
        expect(list1.length()).toEqual(0)
    })

    xit('non-empty list', () => {
        const list1 = new List([1, 2, 3, 4])
        expect(list1.length()).toEqual(4)
    })
})

describe('returns a list of elements whose values equal the list value transformed by the mapping function', () => {
    xit('empty list', () => {
        const list1 = new List()
        expect(list1.map((el: number) => ++el).values).toEqual([])
    })

    xit('non-empty list', () => {
        const list1 = new List([1, 3, 5, 7])
        expect(list1.map((el: number) => ++el).values).toEqual([2, 4, 6, 8])
    })
})

describe('folds (reduces) the given list from the left with a function', () => {
    xit('empty list', () => {
        const list1 = new List()
        expect(list1.foldl((acc: number, el: number) => el / acc, 2)).toEqual(2)
    })

    xit('division of integers', () => {
        const list1 = new List([1, 2, 3, 4])
        expect(list1.foldl((acc: number, el: number) => el / acc, 24)).toEqual(64)
    })
})

describe('folds (reduces) the given list from the right with a function', () => {
    xit('empty list', () => {
        const list1 = new List()
        expect(list1.foldr((acc: number, el: number) => el / acc, 2)).toEqual(2)
    })

    xit('division of integers', () => {
        const list1 = new List([1, 2, 3, 4])
        expect(list1.foldr((acc: number, el: number) => el / acc, 24)).toEqual(9)
    })
})

describe('reverse the elements of a list', () => {
    xit('empty list', () => {
        const list1 = new List()
        expect(list1.reverse().values).toEqual([])
    })

    xit('non-empty list', () => {
        const list1 = new List([1, 3, 5, 7])
        expect(list1.reverse().values).toEqual([7, 5, 3, 1])
    })
})

export default class List<T> {
  values: T[]

  constructor(values: T[] = []) {
    this.values = values
  }

  push(el: T): List<T> {
    return new List([...this.values, el])
  }

  append(list: List<T>): List<T> {
    return list.foldl((acc, el) => acc.push(el), this)
  }

  concat(list: List<List<T>>): List<T> {
    return list.foldl((acc, el) => new List(acc.append(el).values), this)
  }

  filter(fn: (el: T) => boolean): List<T> {
    return this.foldl((acc, el) => (fn(el) && acc.push(el)) || acc, new List())
  }

  length(): number {
    return this.foldl((length) => length + 1, 0)
  }

  map(fn: (el: T) => T): List<T> {
    return this.foldl((acc, el) => acc.push(fn(el)), new List())
  }

  foldl(fn: (acc: any, el: any) => any, init: any): any {
    const [head, ...tail] = this.values
    return head ? new List(tail).foldl(fn, fn(init, head)) : init
  }

  foldr(fn: (acc: T, el: T) => T, init: T): T {
    const [head, ...tail] = this.values
    return head ? fn(new List(tail).foldr(fn, init), head) : init
  }

  reverse(): List<T> {
    return this.foldl((acc, el) => new List([el, ...acc.values]), new List())
  }
}

Community comments

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archanid

Solution Author

commented 214 days ago

I feel pretty good about this solution. The only thing I'm not sure about is whether I should have used .append within .concat, or implemented this:

  concat(list: List<List<T>>): List<T> {
    let newlist: T[] = [...this.values]
    for (const sublist of list.values) {
      newlist = [...newlist, ...sublist]
    }
    return new List(newlist)
  }

I feel really good about the way I did foldl and foldr, with recursion. It took me a little while to understand the order of operations for foldr, though.

All in all, I need to work on problems like this. I'm very slow and it's hard for my brain to think sometimes. I want to improve that.

archanid

Solution Author

commented 214 days ago

Just submitted another iteration. I peeked at another student's solution and my mind was blown. I don't have to pass another list around holding the ...rest of the values. I can just call .foldr and .foldl on a new list containing just the ...rest values! So beautiful and elegant. I can't believe it.

archanid's Reflection

I want to write `foldr` differently, without using `reverse`.

archanid's solution

to List Ops in the TypeScript Track