LeetCode: 1281. Subtract the Product and Sum of Digits of an Integer

Question

I have been trying to figure out this question from LeetCoode. This is the problem: Given an integer number n, return the difference between the product of its digits and the sum of its digits.

This is my code so far:

/**
 * @param {number} n
 * @return {number}
 */
//n=234
var subtractProductAndSum = function(n) {
    var z= n.toString().length;
    var g= n.toString()
    for(var i=0; i<z; i++){
      var p= g[i]+g[i+1]+g[i+2];
    var y=g[i]*g[i+1]*g[i+2];
        var d= y-p
    }
    return d;
};

score 1 · Answer 1 · answered Dec 31 '19 at 17:18

1

Your solution assumes the number has three digits, which is probably not what the problem intended.

I'd run over the number and extract the digits one by one, and sum and multiply them as I go. Then, just subtract the two:

var subtractProductAndSum = function(n) {
    // Initialize the sum and the product with neutral values
    var sum = 0;
    var product = 1;

    while (n > 0) {
        var digit = n % 10;
        sum += digit;
        product *= digit;

        n = Math.floor(n / 10);
    }

    return product - sum;
};

answered Dec 31 '19 at 17:18

Mureinik

297,002
52
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350

Why do you do digit= n % 10; and n = Math.floor(n/10); What's the purpose of this ? – Олег Войтинський Jan 29 '21 at 14:00
1

@ОлегВойтинський `n % 10` returns the remainder of a division by 10, thus returning the last digit. `Math.floor(n / 10)` removes the last digit. – Mureinik Jan 29 '21 at 14:10

score 1 · Answer 2 · answered Dec 31 '19 at 17:19

You can convert the integer into a string and then split it. Then using the array of digit strings, find the product of the digits and the sum of the digits by using the reduce function like so:

const subtractProductAndSum = function(n) {
  const digitArray = n.toString().split('')
  const digitProduct = digitArray.reduce((previousValue, currentValue) => previousValue * parseInt(currentValue), 1)
  const digitSum = digitArray.reduce((previousValue, currentValue) => previousValue + parseInt(currentValue), 0)
  return digitProduct - digitSum
}

score 1 · Answer 3 · answered Dec 31 '19 at 17:25

You can try

/**
 * @param {number} n
 * @return {number}
 */
//n=234
var subtractProductAndSum = function(n) {
let product = (n+"").split("").reduce((total,curr)=>total * +curr,1);
let sum = (n+"").split("").reduce((total,curr)=>total + +curr,0);

return product - sum;
};

or

/**
 * @param {number} n
 * @return {number}
 */
//n=234
var subtractProductAndSum = function(n) {
let  product = 1;
let sum = 0;
n = n+"";

for(let  i = 0; i < n.length; i++){
product *= +n[i];
sum += +n[i];
}

return product - sum;
};

score 0 · Answer 4 · answered Dec 31 '19 at 17:30

See the following :

/**
 * @param {number} n
 * @return {number}
 */
//n=234
var subtractProductAndSum = function(n) {
    var digits = n.split('');
    //console.log(digits);
    var prod = digits.reduce((a, b) => a * parseInt(b), 1);
    //console.log(prod);
    var sum = digits.reduce((a, b) => a + parseInt(b), 0);
    //console.log(sum);
    //console.log(prod - sum);
    return prod - sum;
};

subtractProductAndSum('234');

score 0 · Answer 5 · answered Jan 29 '22 at 18:10

0

//it's working on IntelliJ
public class SubofProSum {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int sum,product,n,value,sub;
        sum = 0;
        product = 1;
        n= in.nextInt();

        while(n>0){
            value=n%10;
            sum = sum+value;
            product = product * value;
            n = n /10;
        }
        System.out.println("sum is: "+sum+" \nMultiple is: "+product);
        sub=product-sum;
        System.out.println("sub is:"+sub);
    }
}

answered Jan 29 '22 at 18:10

Rishav Mehra

1

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Jan 30 '22 at 03:39

score 0 · Answer 6 · edited Jun 14 '22 at 23:28

0

I have solved the same question in Java. I used the modulus operator (%) to compute the last digit, and the code snippet is given below.

public int subtractProductAndSum(int n) {

   int productOfDigits = 1;
    int sumOfDigits = 0;
    while (n > 0) {
        int lastDigit = n % 10;
        productOfDigits *= lastDigit;
        sumOfDigits += lastDigit;
        n /= 10;
    }
    return productOfDigits - sumOfDigits;
}

edited Jun 14 '22 at 23:28

Lee Taylor

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answered Jun 12 '22 at 06:53

Muhammad Mehran

1
1

The question was tagged with javascript, not Java – Corey Cole Jun 12 '22 at 22:21

LeetCode: 1281. Subtract the Product and Sum of Digits of an Integer

6 Answers6