Coding Mock Interview - Product of Array Except for Self (with Ex-Google SWE)

[@tryexponent]

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[@King-ul4nb]

I wish such easy questions were asked in my interviews

[@andrehil]

Correct me if I'm wrong, but no solution presented here had space complexity O(1) as mentioned. That would mean creating no array. Since all solutions created at least an output array, that's O(n). The ones that created more than one were 2n or 3n; some people argue that you can skip the number in that case, which I personally dislike as 2n is twice as much space or time, therefore, not insignificant. The same applies to time when there was more than one for loop.

[@merottm]

Here's a solution with O(1) constant space and O(n) time complexity, written in TypeScript: function productAllExceptI(arr: number[]) { let allProduct = 1 let zerosCount = 0 for (const num of arr) { if (num !== 0) allProduct *= num else zerosCount++ if (zerosCount > 1) break // it'll all be 0 anyway, we can stop multiplying } for (let i = 0; i < arr.length; i++) { if (zerosCount > 1) arr[i] = 0 else if (arr[i] === 0) arr[i] = allProduct else if (zerosCount === 1) arr[i] = 0 else arr[i] = allProduct / arr[i] } return arr }

[@recentvisit]

What about my solution? const getProductArray = (array: number[]): number[] => { const outputArray: number[] = []; for(let i = 0; i < array.length; i ++) { let prod = 1; for(let j = 0; j < array.length; j ++) { if (i !== j) prod*=array[j]; } outputArray.push(prod); } return outputArray }

[@JoeEverest165]

Anytime you have to put a loop inside another loop, take a second to think about some optimization you could do. In this case, why not calculate the product of all items in the array and store it. Then divide by each instance of i

[@vaibhavaaaa]

What IDE are you using for this interview ?

[@tryexponent]

Hey vaibhavaa! It's "CoderPad"

[@cristicode4301]

I don't even understand the question.

[@tankieslayer6927]

This would take me a minute.