# C Exercises: Update every array element with multiplication of previous and next numbers in array

## C Array: Exercise-103 with Solution

Write a program in C to update every array element with multiplication of previous and next numbers in array.

**Sample Solution:**

**C Code:**

```
#include<stdio.h>
// Function to create a new array where each element is the product of its adjacent elements
void newArryPrevNext(int arr1[], int n)
{
if (n <= 1) // If the array has 0 or 1 element, no change is made
return;
int pre_elem = arr1[0]; // Store the first element initially
arr1[0] = arr1[0] * arr1[1]; // Update the first element by multiplying with the next element
// Loop through the array from 2nd to (n-1)th element
for (int i = 1; i < n - 1; i++)
{
int cur_elem = arr1[i]; // Store the current element
arr1[i] = pre_elem * arr1[i + 1]; // Multiply the previous element with the next element and assign to current
pre_elem = cur_elem; // Update the previous element for the next iteration
}
arr1[n - 1] = pre_elem * arr1[n - 1]; // Update the last element of the array by multiplying it with the previous element
}
int main()
{
int arr1[] = {1, 2, 3, 4, 5, 6}; // Given array
int n = sizeof(arr1) / sizeof(arr1[0]); // Calculate the size of the array
int i = 0;
// Print the original array
printf("The given array is: \n");
for(i = 0; i < n; i++)
{
printf("%d ", arr1[i]);
}
printf("\n");
// Process the array to create a new array with elements as the product of adjacent elements
printf("The new array is: \n");
newArryPrevNext(arr1, n);
// Display the elements of the new array
for (int i = 0; i < n; i++)
{
printf("%d ", arr1[i]);
}
return 0;
}
```

Sample Output:

The given array is: 1 2 3 4 5 6 The new array is: 2 3 8 15 24 30

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