C++ Exercises: Update every array element by multiplication of next and previous values of a given array of integers

Last update on December 20 2024 12:26:40 (UTC/GMT +8 hours)

Write a C++ program to update every array element by multiplication of the next and previous values of a given array of integers.

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Sample Solution:

C++ Code :

#include<iostream> // Header file for input/output stream
using namespace std; // Using the standard namespace

// Function to replace each array element by the product of its neighbors
void replace_elements(int nums[], int n)
{
    // If the array has 1 or fewer elements, no replacements can be done
    if (n <= 1)
        return;

    // Initialize the first element with the product of the first two elements
    int prev_element = nums[0];
    nums[0] = nums[0] * nums[1];

    // Loop through the array to replace elements with the product of their neighbors
    for (int i = 1; i < n - 1; i++)
    {
        int curr_element = nums[i];

        // Replace the current element with the product of its neighbors
        nums[i] = prev_element * nums[i + 1];

        // Update the previous element for the next iteration
        prev_element = curr_element;
    }

    // Replace the last element with the product of its neighbor (previous element)
    nums[n - 1] = prev_element * nums[n - 1];
}

int main()
{
    int nums[] = {0, 1, 3, 4, 5, 6, 7, 8, 10}; // Declaration and initialization of an integer array
    int n = sizeof(nums) / sizeof(nums[0]); // Calculate the number of elements in the array
    cout << "Original array: ";
    for (int i = 0; i < n; i++)
        cout << nums[i] << " "; // Output each element of the original array
    replace_elements(nums, n); // Replace array elements with the product of their neighbors
    cout << "\nNew array elements: ";
    for (int i = 0; i < n; i++)
        cout << nums[i] << " "; // Output each element of the modified array
    return 0;
}

Sample Output:

Original array: 0 1 3 4 5 6 7 8 10 
New array elements: 0 0 4 15 24 35 48 70 80 

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