## Multiply numbers represented as arrays in Java?

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I have to write a function that multiplies two numbers represented by two int arrays (so I can't use ArrayLists or something).

Each digit of a number is represented by an `int`

between 0 and 9 in the array, no element should be greater than that.

The first element of the array represents the last digit of the number and so on, therefore the number `1234`

would be `{4,3,2,1}`

as an array in this function.

I thought multiplying those arrays that way would be similar to long multiplication, so I tried to implement it in a similar way: You multiply every digit of the first array with every digit of the second one and store the rest if the result is equal or greater to 10 and then add it to the next digit. However, I seem to have done something wrong in the code (maybe the calculation of the rest??) because the result of my function is not correct: I tested it with 190 times 86 (represented by the arrays `{0,9,1}`

and `{6,8}`

) and get `15342`

(`{2,4,3,5,1}`

) instead of the actual result `16340`

(which would be `{0,4,3,6,1}`

).

Can somebody here help me out with this please? This is my code:

import java.util.Arrays; public class MultiplyArrays { static int[ ] times(int[ ] a, int[ ] b) { int[] arr = new int[a.length + b.length - 1];//arr should be the result of a*b. The result shouldn't be shorter than that int tmp = 0;//stores the rest of two-digit numbers for(int i = b.length - 1; i >= 0; i--){ for(int j = 0; j < a.length; j++){//should multiply all digits of a with the current index of b arr[i + j] = (arr[i + j] + (b[i] * a[j] + tmp)) % 10;//sets the value of the (i+j)th index in arr to the multiplication of two numbers from a and b adding the rest tmp. if((arr[i + j] + b[i] * a[j] + tmp) < 10){//if this number is less than 10, there is no rest tmp = 0; } else{//otherwise, the rest should be the second digit tmp = (((arr[i + j] + (b[i] * a[j] + tmp))) - ((arr[i + j] + (b[i] * a[j] + tmp)) % 10)) / 10;//something in the formula for the rest is wrong, I guess } } } if(tmp != 0){//if the last number of the array containing the result is calculated and there still is a rest, a new array with one more digit is created int[] arr2 = new int[arr.length + 1]; for(int i = arr.length - 1; i >= 0; i--){//the new array copies all numbers from the old array arr2[i] = arr[i]; arr2[arr2.length - 1] = tmp;//the last space is the rest now } return arr2; } else{//if there is no rest after calculating the last number of arr, a new array isn't needed return arr; } } public static void main(String[] args) {//test the function with 190 * 86 int[] a = {0,9,1}; int[] b = {6,8}; System.out.println(Arrays.toString(times(a,b))); } }

Maybe this comes from the fact that your indices in the for-loops of the times()-method are incrementing AND decrementing. The i is going down and the j is going up. Also, in the second for loop, you should only increment to 'a.length - 1', not to 'a.length'.

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Arbitrary precision multiplication is more complex than it seems, and contains corner cases (like one and zero). Fortunately, Java has an arbitrary precision type; `BigInteger`

. In order to use it here, you would need to create two additional methods; one for converting an `int[]`

to a `BigInteger`

, and the second the convert a `BigInteger`

to an `int[]`

.

The first can be done with a single loop adding each digit at index `i`

(multiplied by 10i) to a running total. Like,

private static BigInteger fromArray(int[] arr) { BigInteger bi = BigInteger.ZERO; for (int i = 0, pow = 1; i < arr.length; pow *= 10, i++) { bi = bi.add(BigInteger.valueOf(arr[i] * pow)); } return bi; }

And the second can be done a number of ways, but the easiest is simply to convert the `BigInteger`

to a `String`

to get the `length()`

- once you've done that, you know the length of the output array - and can populate the digits in it. Like,

private static int[] toArray(BigInteger bi) { String s = bi.toString(); int len = s.length(); int[] r = new int[len]; for (int i = 0; i < len; i++) { r[i] = s.charAt(len - i - 1) - '0'; } return r; }

Finally, call those two methods and let `BigInteger`

perform the multiplication. Like,

static int[] times(int[] a, int[] b) { BigInteger ba = fromArray(a), bb = fromArray(b); return toArray(ba.multiply(bb)); }

Running your original `main`

with those changes outputs (as expected)

[0, 4, 3, 6, 1]

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**Java exercises: Multiply corresponding elements of two arrays of ,** Java programming exercises and solution: Write a Java program to multiply corresponding elements of two arrays of integers. The following Java program demonstrates multiplication of two numbers. This is a trivial example program, however it shows how to handle input and output, performing calculations and the use of library classes such as Scanner in Java language. The user is prompted to enter first and the second number.

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**Java Program to Multiply Two Numbers,** When you start learning java programming, you get these type of problems in your assignment. Here we will see two Java programs, first program takes two. In this article, we will write a program to multiply (or divide) every element of an array with a constant number and print the output. Submitted by Abhishek Pathak , on October 15, 2017 JavaScript is the language of the web, but with Node.js this scenario has changed.

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##### Comments

- Is this a homework question? I'd convert it into a string, reverse it, convert to int, multiply both, then re-convert back. A better way could be to sum each number, increasing by
`10**n`

every number (and looping through array backwards) then multiplying the resulting two integers.