C Exercises: Find the product of an array of specified condition

C Array: Exercise-65 with Solution

Write a program in C to find the product of an array such that product is equal to the product of all the elements of arr[] except arr[i].

Expected Output :
The given array is : 1 2 3 4 5 6
The product array is: 720 360 240 180 144 120

The objective is to write a C program that calculates the product array, where each element in the product array is the product of all elements in the original array except the element at the same index. The program should efficiently compute this product without using division to ensure accuracy and handle cases where elements might be zero.

Visual Presentation:

Sample Solution:

C Code:

#include <stdio.h>

void productOfArray(int arr1[], int n) {
    // Arrays to store left and right products of elements
int l_arr[n], r_arr[n], product[n];
int i, j;

    // Initialize the left-most element of left product array
    l_arr[0] = 1;

    // Initialize the right-most element of right product array
    r_arr[n - 1] = 1;

    // Calculate the left product of each element in the array
for (i = 1; i < n; i++)
        l_arr[i] = arr1[i - 1] * l_arr[i - 1];

    // Calculate the right product of each element in the array
for (j = n - 2; j >= 0; j--)
        r_arr[j] = arr1[j + 1] * r_arr[j + 1];

    // Calculate the final product array by multiplying left and right products
for (i = 0; i < n; i++)
product[i] = l_arr[i] * r_arr[i];

    // Print the product array
for (i = 0; i < n; i++)
printf("%d ", product[i]);
}

int main() {
int arr1[] = {1, 2, 3, 4, 5, 6};
int n = sizeof(arr1) / sizeof(arr1[0]);
int i;

    // Print original array
printf("The given array is :  ");
for (i = 0; i < n; i++) {
printf("%d  ", arr1[i]);
    } 
printf("\n");

    // Calculate and print the product array
printf("The product array is: ");
productOfArray(arr1, n);
return 0;
}

Output:

The given array is :  1  2  3  4  5  6  
The product array is: 720 360 240 180 144 120 

Flowchart:

C Programming Code Editor:

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