Sum of Two Arrays - Python

Question

Two random integer arrays/lists have been given as ARR1 and ARR2 of size N and M respectively. Both the arrays/lists contain numbers from 0 to 9(i.e. single digit integer is present at every index). The idea here is to represent each array/list as an integer in itself of digits N and M. You need to find the sum of both the input arrays/list treating them as two integers and put the result in another array/list i.e. output array/list will also contain only single digit at every index.

NOTE: The sizes N and M can be different.

Output array/list(of all 0s) has been provided as a function argument. Its size will always be one more than the size of the bigger array/list. Place 0 at the 0th index if there is no carry.

No need to print the elements of the output array/list.

def sumOfTwoArrays(arr1, n, arr2, m, output) :
    #Your code goes here






#Taking Input Using Fast I/O
def takeInput() :
    n = int(stdin.readline().rstrip())
    if n == 0 :
        return list(), 0
    
    arr = list(map(int, stdin.readline().rstrip().split(" ")))
    return arr, n


#to print the array/list
def printList(arr, n) :
    for i in range(n) :
        print(arr[i], end = " ")
    
    print()


#main
t = int(stdin.readline().rstrip())

while t > 0 :
    arr1, n = takeInput()
    arr2, m = takeInput()
    
    outputSize = (1 + max(n, m))
    output = outputSize * [0]
    
    sumOfTwoArrays(arr1, n, arr2, m, output)
    printList(output, outputSize)
    
    t -= 1

Sample Input: 1

3

6 2 4

3

7 5 6

Sample Output: 1 3 8 0

Answer 1

This problem can be solved by a simple function like this: (Note - you can made adjustment on the last line of return result if needed to meet the "strange requirement" of - 'place 0 at the 0th index if there is no carry'. It's left as a trivial exercise. )

def sum_two_array(L1, L2):
    carry, total = 0, 0
    m, n = len(L1), len(L2)
    k   = max(m, n)
    
    result = [0] + [0] * k  # add +1 

    for i in range(1, k+1):
        a = L1[m-i] if m - i >= 0 else 0
        b = L2[n-i] if n - i >= 0 else 0
     
        total = a + b + carry
        result[k-i + 1] = total % 10
        carry = total // 10
    
    if carry > 0: result[0] = carry
    
    return result if result[0] != 0 else result[1:]

if __name__ == '__main__':
    L1 = [6, 4, 4]
    L2 = [7, 5, 6]

    print(sum_two_array(L1, L2))  # [1, 4, 0, 0]

    print(sum_two_array([6, 2, 4], [7, 5, 6]))  # [1, 3, 8, 0]

    print(sum_two_array([1, 2, 4], [8, 0]))     # [2, 0, 4]

Answer 2

JAVA CODE:

import java.lang.Math;

public class Solution {

public static void sumOfTwoArrays(int arr1[], int arr2[], int output[]) {
    
    int n = arr1.length; //size of arr1
    int m = arr2.length; //size of arr2
    int o = n+1; //size of output array
    int sum1 = 0, sum2=0, totalSum=0;
    
    //storing sum of arr1 in sum1 
    for(int i=0; i<n; i++)
    {
        sum1+= arr1[i] * Math.pow(10,(n-1-i));
    }
    
    //storing sum of arr2 in sum2
    for(int i=0; i<m; i++)
    {
        sum2+= arr2[i] * Math.pow(10, (m-1-i));
    }
    
    totalSum = sum1+sum2;
    
    //storing totalSum in reverse order in output array
    for(int i=o-1; i>=0; i--)
    {
        output[i] = totalSum % 10;
        totalSum = totalSum/10;
    }
    
    
}

}

Explanation: condition: arr1[n], arr2[m], output[n+1]

1. Instead of calculating sum of unit, tens, and so on digits of both the arrays.
2. we first calculate the sum1 of arr1, and sum2 of arr2, by using: number at index * (10 ^ ((n-1) - index)) concept.
3. sum1 and sum2 are equal to the n and m sized numbers of respective arrays
4. we store totalSum = sum1+sum2
5. we store totalSum's every digit in output[n+1] array
6. we store it in reverse order by using % and / operations