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Find the maximum sum of digits of the product of two numbers

  • Last Updated : 16 Mar, 2021

Given an array arr[] of size N( > 2). The task is to find the maximum sum of digits of the product of any two numbers of the given array.
Examples: 
 

Input : arr[] = {8, 7} 
Output : 11 
The product of 8 and 7 is 56. The sum of the digits of 56 is equal to 11.
Input : arr[] = {4, 3, 5} 
Output :
Product of 4 & 3 = 12. Sum of the digits = 3. 
Product of 3 & 5 = 15. Sum of the digits = 6. 
Product of 4 & 5 = 20. Sum of the digits = 2.
 

 

Approach: Run nested loops to select two numbers of the array and get the product. For every product check the digit sum and find the maximum digit sum.
Below is the implementation of the above approach: 
 

C++




// C++ program find the maximum sum of
// digits of the product of two numbers
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sum of the digits
int sumDigits(int n)
{
    int digit_sum = 0;
    while (n) {
        digit_sum += n % 10;
        n /= 10;
    }
    return digit_sum;
}
 
// Function to find the maximum sum of digits of product
int productOfNumbers(int arr[], int n)
{
    int sum = INT_MIN;
 
    // Run nested loops
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
            int product = arr[i] * arr[j];
 
            // Find the maximum sum
            sum = max(sum, sumDigits(product));
        }
    }
 
    // Return the required answer
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 4, 3, 5 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << productOfNumbers(arr, n);
 
    return 0;
}

Java




// Java program find the maximum sum of
// digits of the product of two numbers
import java.io.*;
 
class GFG
{
 
// Function to find the sum of the digits
static int sumDigits(int n)
{
    int digit_sum = 0;
    while (n > 0)
    {
        digit_sum += n % 10;
        n /= 10;
    }
    return digit_sum;
}
 
// Function to find the maximum sum
// of digits of product
static int productOfNumbers(int []arr, int n)
{
    int sum = Integer.MIN_VALUE;
 
    // Run nested loops
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            int product = arr[i] * arr[j];
 
            // Find the maximum sum
            sum = Math.max(sum, sumDigits(product));
        }
    }
 
    // Return the required answer
    return sum;
}
 
// Driver code
public static void main (String[] args)
{
    int []arr = { 4, 3, 5 };
     
    int n = arr.length;
     
    System.out.print( productOfNumbers(arr, n));
}
}
 
// This code is contributed by anuj_67..

Python3




# Python3 program find the maximum sum of
# digits of the product of two numbers
import sys
 
# Function to find the sum of the digits
def sumDigits(n):
 
    digit_sum = 0;
    while (n > 0):
        digit_sum += n % 10;
        n /= 10;
     
    return digit_sum;
 
# Function to find the maximum sum
# of digits of product
def productOfNumbers(arr, n):
 
    sum = -sys.maxsize - 1;
 
    # Run nested loops
    for i in range(n - 1):
        for j in range(i + 1, n):
            product = arr[i] * arr[j];
 
            # Find the maximum sum
            sum = max(sum, sumDigits(product));
 
    # Return the required answer
    return sum;
 
# Driver code
if __name__ == '__main__':
 
    arr =[ 4, 3, 5 ];
 
    n = len(arr);
 
    print(int(productOfNumbers(arr, n)));
 
# This code contributed by PrinciRaj1992

C#




// C# program find the maximum sum of
// digits of the product of two numbers
using System;
 
class GFG
{
 
// Function to find the sum of the digits
static int sumDigits(int n)
{
    int digit_sum = 0;
    while (n > 0)
    {
        digit_sum += n % 10;
        n /= 10;
    }
    return digit_sum;
}
 
// Function to find the maximum sum
// of digits of product
static int productOfNumbers(int []arr, int n)
{
    int sum = int.MinValue;
 
    // Run nested loops
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            int product = arr[i] * arr[j];
 
            // Find the maximum sum
            sum = Math.Max(sum, sumDigits(product));
        }
    }
 
    // Return the required answer
    return sum;
}
 
// Driver code
public static void Main (String[] args)
{
    int []arr = { 4, 3, 5 };
     
    int n = arr.Length;
     
    Console.Write(productOfNumbers(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
    // Javascript program find the maximum sum of
    // digits of the product of two numbers
     
    // Function to find the sum of the digits
    function sumDigits(n)
    {
        let digit_sum = 0;
        while (n > 0) {
            digit_sum += n % 10;
            n = parseInt(n / 10, 10);
        }
        return digit_sum;
    }
 
    // Function to find the maximum sum of digits of product
    function productOfNumbers(arr, n)
    {
        let sum = Number.MIN_VALUE;
 
        // Run nested loops
        for (let i = 0; i < n - 1; i++) {
            for (let j = i + 1; j < n; j++) {
                let product = arr[i] * arr[j];
 
                // Find the maximum sum
                sum = Math.max(sum, sumDigits(product));
            }
        }
 
        // Return the required answer
        return sum;
    }
       
    let arr = [ 4, 3, 5 ];
   
    let n = arr.length;
   
    document.write(productOfNumbers(arr, n));
     
</script>

Output: 
 



6

 

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