Daily Coding Day Two

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Round two of daily coding challenges. This one was marked as hard in the email, and I am sure that there is a way to do this by extending reduce or perhaps filter to encompass the for loop, but I am interested more in getting a solution to these exercises rather than the solution. Day two here it is:

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// Given an array of integers, return a new array such that each element 
// at index i of the new array is the product of all the numbers in 
// the original array except the one at i.

// For example, if our input was [1, 2, 3, 4, 5], the expected output 
// would be [120, 60, 40, 30, 24]. If our input was [3, 2, 1], 
// the expected output would be [2, 3, 6]

function productOfNumbers(numbers)
{    
    let toReturn = []

    for (let i = 0; i < numbers.length; i++) 
    {
        // Grab the element    
        const n = numbers[i];

        // Add the multipled values to the final array
        toReturn.push(numbers.filter(x => x != n).reduce((a,b) => a*b, 1))
    }

    console.log(toReturn)
    return toReturn
}