Question

109

Determining if an integer can be expressed as a palindromic sum

rated 0 times [ 110] [ 1] / answers: 1 / hits: 15641 / 5 Years ago, thu, february 14, 2019, 12:00:00

I was asked this question in an interview:

An integer is special if it can be expressed as a sum that's a palindrome (the same backwards as forwards). For example, 22 and 121 are both special, because 22 equals 11+11 and 121 equals 29+92.

Given an array of integers, count how many of its elements are special.

but I couldn't think of any solution. How can this be done?

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freddiej

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2 Years ago

answered 5 Years ago collinisaaka · Accepted Answer

In the stress and the hurry of an interview, I would have certainly found a dumb and naive solution.

pseudo code

loop that array containing the numbers

    Looping from nb = 0 to (*the number to test* / 2)

        convert nb to string and reverse the order of that string (ie : if you get 29, transform it to 92)

        convert back the string to a nb2

        if (nb + nb2 == *the number to test*)

            this number is special. Store it in the result array

    end loop

end loop

print the result array

function IsNumberSpecial(input)

{

    for (let nb1 = 0; nb1 <= (input / 2); ++nb1)

    {

        let nb2 = parseInt(( + nb1).split().reverse().join()); // get the reverse number

        if (nb2 + nb1 == input)

        {

           console.log(nb1 +  +  + nb2 +  =  + input);

            return (true);

        }

    }

    return (false);

}



let arr = [22, 121, 42];



let len = arr.length;

let result = 0;



for (let i = 0; i < len; ++i)

{

    if (IsNumberSpecial(arr[i]))

        ++result;

}



console.log(result +  number + ((result > 1) ? s : ) +  found);