Typescript: Working with objects and using for...in

Question

I converted the following function from JS to Typescript and it works well. However in the line if (!(key * key) in frequencyCounter2) { i get the error: "The left-hand side of an 'in' expression must be of type 'any', 'string', 'number', or 'symbol'."

I tried to typecast key as a number, but to no avail. Is this operation generally not senseful when using typescript?

// The function same accepts two arrays and should return true if every value in the array has its corresponding 
// value squared in the second array. The frequency of values must be the same.
function sameOn(arrA: number[], arrB: number[]): boolean {
    // complexity O(n): If arr is 1000 lines, this runs 2000 times
    if (arrA.length !== arrB.length) {
        return false;
    }
    type CountType = {
        [key: number] : number
    }

    const frequencyCounter1: CountType = {};
    const frequencyCounter2: CountType = {};

    for (const val of arrA) {
        frequencyCounter1[val] = (frequencyCounter1[val] || 0) +1;
    }
    for (const val of arrB) {
        frequencyCounter2[val] = (frequencyCounter2[val] || 0) +1;
    }

    for (const key in frequencyCounter1) {               
        if (!(key * key) in frequencyCounter2) {
            return false;
        }
        if (frequencyCounter2[key * key] !== frequencyCounter1[key]) {
            return false;
        }
    }
    return true;
}

sameOn([1,2,3,2], [4,1,9,4]) // returns true

T.J. Crowder · Accepted Answer · 2020-08-19 08:48:05Z

4

This line:

if (!(key * key) in frequencyCounter2) {

looks for a boolean in frequenceCounter2, because key * key creates a number and then !number converts that number to boolean.

If you're trying to check that the key * key key isn't there, you need more ():

if (!((key * key) in frequencyCounter2)) {
//   ^−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−^−−−−−−−−−

or really, you just need to relocate your first ):

if (!(key * key in frequencyCounter2)) {
// Not here −−−^−−−−−−−−−−−−−−−−−−−−^−− here

since the key * key part of the in expression will be evaluated before the in part.

It sounds like this may well have been a bug in the original JavaScript, which would have checked for "true" or "false" in frequencyCounter2 without raising any complaints. :-)

edited Aug 19, 2020 at 8:48

answered Aug 19, 2020 at 7:05

T.J. Crowder

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malte238749874 · Accepted Answer · 2020-08-20 07:11:40Z

The refactored solution is:

export function sameOn(arrA: number[], arrB: number[]): boolean {
    // complexity O(n): If arr is 1000 lines, this runs 2000 times
    if (arrA.length !== arrB.length) {
        return false;
    }
    type CountType = {
        [key: number] : number
    }

    const frequencyCounter1: CountType = {};
    const frequencyCounter2: CountType = {};

    for (const val of arrA) {
        frequencyCounter1[val] = (frequencyCounter1[val] || 0) +1;
    }
    for (const val of arrB) {
        frequencyCounter2[val] = (frequencyCounter2[val] || 0) +1;
    }
    
    for (const key in frequencyCounter1) {
        const squaredKey = Number(key)*Number(key);
                  
        if (!(String(squaredKey) in frequencyCounter2)) {
            return false;
        }
        if (frequencyCounter2[squaredKey] !== frequencyCounter1[key]) {
            return false;
        }
    }
    return true;
}

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