Given an array arr, the task is to find the sum of the product of digits of all array elements
Example:
Input: arr[]={11, 23, 41}
Output: 11
Explanation: 1*1 + 2*3 + 4*1 = 1 + 6 + 4 = 1111Input: arr[]={46, 32, 78, 0}
Output: 86
Approach: To solve this problem, find the product of digits of all numbers and then just add them up. Follow the below steps to solve this problem:
- Create a function findProduct which will get a number and find the product of its digits.
- Create a variable sum to store the final answer and initialise it with 0.
- Now, traverse the array and for each element:
- Pass it to the function findProduct and get the product of its digit.
- Then add its product of digit to sum.
- Return sum as the final answer.
Below is the implementation of the above approach:
C++
// C++ code for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the product of the // digits of a number N int findProduct(int N) { if (N == 0) { return 0; } int product = 1; while (N > 0) { product *= (N % 10); N /= 10; } return product; } // Function to find the sum of the product of // digits of all array elements int sumOfProduct(vector<int> arr) { int sum = 0; for (auto x : arr) { sum += findProduct(x); } return sum; } // Driver Code int main() { vector<int> arr = { 46, 32, 78, 0 }; cout << sumOfProduct(arr); } |
Java
// Java code to implement above approach import java.util.*; public class GFG { // Function to find the product of the // digits of a number N static int findProduct(int N) { if (N == 0) { return 0; } int product = 1; while (N > 0) { product *= (N % 10); N /= 10; } return product; } // Function to find the sum of the product of // digits of all array elements static int sumOfProduct(int []arr) { int sum = 0; for (int i = 0; i < arr.length; i++) { sum += findProduct(arr[i]); } return sum; } // Driver code public static void main(String args[]) { int []arr = { 46, 32, 78, 0 }; System.out.println(sumOfProduct(arr)); } } // This code is contributed by Samim Hossain Mondal. |
Python3
# Python code for the above approach # Function to find the product of the # digits of a number N def findProduct(N): if (N == 0): return 0 product = 1 while (N > 0): product *= (N % 10) N = N // 10 return product # Function to find the sum of the product of # digits of all array elements def sumOfProduct(arr): sum = 0 for x in arr: sum += findProduct(x) return sum # Driver Code arr = [46, 32, 78, 0] print(sumOfProduct(arr)) # This code is contributed by Saurabh Jaiswal |
C#
// C# code to implement above approach using System; class GFG { // Function to find the product of the // digits of a number N static int findProduct(int N) { if (N == 0) { return 0; } int product = 1; while (N > 0) { product *= (N % 10); N /= 10; } return product; } // Function to find the sum of the product of // digits of all array elements static int sumOfProduct(int []arr) { int sum = 0; for (int i = 0; i < arr.Length; i++) { sum += findProduct(arr[i]); } return sum; } // Driver code public static void Main() { int []arr = { 46, 32, 78, 0 }; Console.Write(sumOfProduct(arr)); } } // This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript code for the above approach // Function to find the product of the // digits of a number N function findProduct(N) { if (N == 0) { return 0; } let product = 1; while (N > 0) { product *= (N % 10); N = Math.floor(N / 10) } return product; } // Function to find the sum of the product of // digits of all array elements function sumOfProduct(arr) { let sum = 0; for (let x of arr) { sum += findProduct(x); } return sum; } // Driver Code let arr = [46, 32, 78, 0]; document.write(sumOfProduct(arr)); // This code is contributed by Potta Lokesh </script> |
Output:
86
Time Complexity: O(NlogM), where N is the size of the array and M is the maximum number in the array
Auxiliary Space: O(1)
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