Java: How to sum the elements of two arrays with different lengths

Question

I am trying to add the elements of two arrays with different lengths together. The code below is only for the same length and here is all I have so far.

    //for the same lengths
int[]num1 = {1,9,9,9};
int[]num2 = {7,9,9,9};// {9,9,9}

    int total = 0, carry = 1;
    int capacity = Math.max(num1.length,num2.length);
    int []arraySum = new int [capacity];

        for (int i = capacity - 1 ; i >= 0; i--)
        {   
            arraySum[i] = num1[i]+ num2[i];

            if (arraySum[i] > 9)
            {
                arraySum[i] = arraySum[i] % 10;

                num2[i-1] = num2[i-1] + carry;  
            }


        }

    for(int i = 0; i < arraySum.length; i++)
    {
        System.out.print(arraySum[i]);
    }

What should I do if I change the elements in num2 and length to like {9,9,9}? I know I probably need to put another for-loop as an inside for-loop and control the indices of the array with smaller length but how....?? One more thing... what should I do for those for-loops conditions because num1 and num2 will be eventually INPUTED by the user.

Well, you can tell that the inputs are limited because if num1[0] + num2[0] > 9 the carry has no index to be placed, then it can't be compiled. So, I need to shift the whole array to the right and place the carry from num1[0] + num2[0]. Here is the problem!! Where should I put the shifting code? I am kinda confused.......

Don't mix god with programming. The easiest way is to check out which array is shorter, and copy that array to a same-sized array as the bigger one, but pad it with zeroes. — Kayaman, Nov 17 '13 at 08:44
Start with an array of size capacity + 1 and then you can place the carry in the zeroth place. — Chandranshu, Nov 17 '13 at 08:50
Everything would be much easier if you stored the digits in reverse order (so the first digit in each array is the least significant). Then you could use the answer by JB Nizet. — Ted Hopp, Nov 17 '13 at 08:58
Actually, you declare an int[] array with the capacity as Math.max(num1.length, num2.length). It is not encough. You should set the capacity as Math.max(num1.length, num2.length) +1. Why? See if num1 is {1,9,9,9} and num2 is {9,9,9,9}, how can the arraySum to represent the sum {1,1,9,9,8}? So we need to declare it as below to consider if carry is needed. int[] arraySum = new int[capacity + 1]; — Mengjun, Nov 17 '13 at 10:13
you can convert the two arrays into two numbers, add them, then break it apart into a resultant folder. — Angel Koh, Feb 11 '18 at 12:11

JB Nizet · Answer 1 · 2013-11-17T09:01:03.207

2

It's quite simpe actually. Inside your loop, check that the current index is valid for both array, and replace the value to add by 0 in case of an invalid index:

int value1 = (i < num1.length) ? num1[i] : 0;
int value2 = (i < num2.length) ? num2[i] : 0;
arraySum[i] = value1 + value2;

EDIT: I didn't understand that you wanted to right-align the arrays and not left-align them. The simplest solution is probably to write and read everything in the arrays in reverse order. So if the numbers are 456 and 7658, the first array would contain 6, 5, 4 and the second one would contain 8, 5, 6, 7.

edited Nov 17 '13 at 09:01

answered Nov 17 '13 at 08:51

JB Nizet

678,734
91
1,224
1,255

@TedHopp: I edited my answer and suggested the same trick as in your comment (that you added while I was editing my answer :-)) – JB Nizet Nov 17 '13 at 09:01

Mengjun · Accepted Answer · 2013-11-17T11:11:55.440

Actually, you declare an int[] array with the capacity as Math.max(num1.length, num2.length).

It is not encough. You should set the capacity as Math.max(num1.length, num2.length) +1.

Why?

See if num1 is {1,9,9,9} and num2 is {9,9,9,9}, how can the arraySum to represent the sum {1,1,9,9,8}?

So we need to declare it as below to consider if carry is needed.

 int[] arraySum = new int[capacity + 1];

Then when print the sum, check if arraySum[0] is 0 or 1, if it euqals to 0, do not print it in Console.

Modified code for reference is as follows:

package question;

public class Example {

public static void main(String[] args) {

    // for the same lengths
    int[] num1 = { 1,9,9,9 };
    int[] num2 = { 9,9,9,9};// {9,9,9}

    // 1999+9999 = 11998, its length is greater than the max
    int capacity = Math.max(num1.length, num2.length);
    int[] arraySum = new int[capacity + 1];

    int len2 = num2.length;
    int len1 = num1.length;

    if (len1 < len2) {
        int lengthDiff = len2 - len1;

        /*
         * Flag for checking if carry is needed.
         */
        boolean needCarry = false;

        for (int i = len1 - 1; i >= 0; i--) {
            /**
             * Start with the biggest index
             */
            int sumPerPosition =0;

            if (needCarry) {
                 sumPerPosition = num1[i] + num2[i + lengthDiff] +1;
                 needCarry = false;
            }else
            {
                 sumPerPosition = num1[i] + num2[i + lengthDiff];
            }

            if (sumPerPosition > 9) {
                    arraySum[i + lengthDiff + 1] = sumPerPosition % 10;
                needCarry = true;
            }else
            {
                arraySum[i + lengthDiff + 1] = sumPerPosition % 10;
            }
        }

        /**
         * Handle the remaining part in nun2 Array
         */

        for (int i = lengthDiff - 1; i >= 0; i--) {
            /*
             * Do not need to care num1 Array Here now
             */

            if(needCarry){
                arraySum[i + 1] = num2[i]+1;
            }else
            {
                arraySum[i + 1] = num1[i] ;
            }

            if (arraySum[i + 1] > 9) {
                arraySum[i + 1] = arraySum[i + 1] % 10;
                needCarry = true;
            } else {
                needCarry = false;
            }


        }

        /*
         * Handle the last number, if carry is needed. set it to 1, else set
         * it to 0
         */
        if (needCarry) {
            arraySum[0] = 1;
        } else {
            arraySum[0] = 0;
        }

    } else {
        int lengthDiff = len1 - len2;

        /*
         * Flag for checking if carry is needed.
         */
        boolean needCarry = false;

        for (int i = len2 - 1; i >= 0; i--) {
            /**
             * Start with the biggest index
             */
            int sumPerPosition = 0;


            if (needCarry) {
                 sumPerPosition = num2[i] + num1[i + lengthDiff] +1;
                 needCarry = false;
            }else
            {
                 sumPerPosition = num2[i] + num1[i + lengthDiff];
            }

            if (sumPerPosition > 9) {
                    arraySum[i + lengthDiff + 1] = sumPerPosition % 10;
                needCarry = true;
            }else
            {
                arraySum[i + lengthDiff + 1] = sumPerPosition % 10;
            }
        }

        /**
         * Handle the remaining part in nun2 Array
         */

        for (int i = lengthDiff - 1; i >= 0; i--) {
            /*
             * Do not need to care num1 Array Here now
             */

            if(needCarry){
                arraySum[i + 1] = num1[i]+1;
            }else
            {
                arraySum[i + 1] = num1[i] ;
            }

            if (arraySum[i + 1] > 9) {
                arraySum[i + 1] = arraySum[i + 1] % 10;
                needCarry = true;
            } else {
                needCarry = false;
            }

        }

        /*
         * Handle the last number, if carry is needed. set it to 1, else set
         * it to 0
         */
        if (needCarry) {
            arraySum[0] = 1;
        } else {
            arraySum[0] = 0;
        }
    }

    /*
     * Print sum 
     * 
     * if arraySum[0] ==1, print 1
     * 
     * Do not print 0 when arraySum[0] ==0
     */
    if(arraySum[0] == 1)
    {
        System.out.print(1);
    }
    for (int i = 1; i < arraySum.length; i++) {

        System.out.print(arraySum[i]);
    }
}
}

An example that num1 is {1,9,9,9} and num2 is {9,9,9,9}, the sum result is as follows:

output in Console:

There is something wrong with the carry. If I change the arrays' lengths and their elements, the output is wrong but it is pretty close. Let's say, num1{1,2} + num2{1,1,0}. The output should be 121, but it becomes 100... — coding, Nov 17 '13 at 10:46
Updated the code. In your example num1{1,2} + num2{1,1,0}. Its output should be 122. :) — Mengjun, Nov 17 '13 at 11:13

score 0 · Answer 3 · answered Nov 17 '13 at 09:17

Here is more performance centric solution:

public class SumArrays {

    public static void main(String[] args) {

        int[] num1 = {1, 9, 9, 9};
        int[] num2 = {7, 9, 9, 9, 9, 9, 9};

        int[] biggerArray = num1.length > num2.length ? num1 : num2;
        int[] smallerArray = num1.length <= num2.length ? num1 : num2;

        int[] summedArray = new int[biggerArray.length];
        System.arraycopy(biggerArray, 0, summedArray, 0, biggerArray.length);

        for (int i = 0; i < smallerArray.length; i++) {
            summedArray[i] += smallerArray[i];
        }

        for (int i = 0; i < summedArray.length; i++) {
            System.out.println(summedArray[i]);
        }
    }
}

this might not work if his output int array requires the numbers to be 0-9 — Greg, Nov 17 '13 at 09:20
My simple solution only sums the elements of two arrays with different lengths (As it's described in title). However it could be customized to meeting some other custom requirement — Nikoloz, Nov 17 '13 at 09:40

score 0 · Answer 4 · edited Feb 11 '18 at 12:06

You can use an ArrayList instead of an array and implement a normal addition method to it:

public static ArrayList<Integer> sum(int[] arr, int[] arr2) {
    ArrayList<Integer> al = new ArrayList<>();
    int i = arr.length - 1;
    int j = arr2.length - 1;
    int c = 0;
    while (i >= 0 && j >= 0) {
        int temp = arr[i] + arr2[j] + c;
        if (temp >= 10) {
            int r = temp % 10;
            al.add(0, r);
            c = temp / 10;
        } else {
            al.add(0, temp);
            c = 0;
        }
        i--;
        j--;
    }
    if (i < 0 && j >= 0) {
        while (j >= 0) {
            al.add(0, arr2[j] + c);
            c = 0;
            j--;
        }
    } else if (j < 0 && i >= 0) {
        while (i >= 0) {
            al.add(0, arr[i] + c);
            c = 0;
            i--;
        }

    } else
        al.add(0, c);
    return al;
}

score -1 · Answer 5 · answered Nov 17 '13 at 09:14

for kicks, here is an alternative:

int[] num1 =
{ 1, 9, 9, 9 };
int[] num2 =
{ 7, 9, 9, 9 };

//convert the int array to a string
StringBuilder sb = new StringBuilder(num1.length);
for (int i : num1)
{
    sb.append(i);
}
String sNum1 = sb.toString();

System.out.println(sNum1);

StringBuilder sb2 = new StringBuilder(num2.length);
for (int i : num2)
{
    sb2.append(i);
}
String sNum2 = sb2.toString();

System.out.println(sNum2);

try
{
    //parse the string to an int
    int iNum1 = Integer.parseInt(sNum1);
    int iNum2 = Integer.parseInt(sNum2);

    //add them together
    int sum = iNum1 + iNum2;
    String sSum = Integer.toString(sum);

    System.out.println(sSum);

    // convert num back to array
    int[] sumArray = new int[sSum.length()];
    for (int i = 0; i < sSum.length(); i++)
    {
        sumArray[i] = sSum.charAt(i) - '0';
        System.out.println(sumArray[i]);
    }
}
catch (Exception e)
{
    // couldnt parse ints
}

Java: How to sum the elements of two arrays with different lengths

5 Answers5

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