# Replace each element of array by multiplication of previous and next in C++

From the title of the problem, it is clear that for a given array, we have to Replace each element by multiplication of previous and next element. Means for each index number **i, **the **arr[i]** will be equal to multiplication of **arr[i – 1]** and **arr[i + 1].**

You may ask a question that what about the first and last element of the array as for the first element there is no previous element and for the last element, there is no next element. well, for them we can multiply that element with previous or next element depending on the availability. That is, for the first element **arr[0]**, we can multiply the first and next element **(arr[0] * arr[1]) **and save it to **arr[0].** and for the last element **arr[n]**, we can multiply the last element and previous element **(arr[n – 1] * arr[n]) **save to **arr[n]**.

Let’s take an example to understand the problem in a better way.

**Example: **

```
input array : arr[] = {2, 5, 8, 6, 7, 3, 4}
Output array will be : arr[] = {2*5, 2*8, 5*6, 8*7, 6*3, 7*4, 3*4}
Therefore,
```**arr[] = {10, 16, 30, 56, 18, 28, 12}**

## Algorithm to replace each element of an array by multiplication of previous and next

- Declare and initialize an array.
- For the first element multiply
**arr[0]**and**arr[1]**and save to**arr[0].** - For the last element multiply
**arr[n – 1]**and**arr[n – 2]**and save to**arr[n – 1]** - declare a variable for the previous element. Consider
**previous.** - For rest elements, start iteration from
**i = 0 to size of the array.**Save each arr[i] to a variable for the current reference. consider**current.** - multiply the
**previous**and**arr[i + 1]**and save to**arr[i].** - update
**previous**by**current**. - end the iteration
- print the elements of modified array.

## C++ program to replace each element of an array by multiplication of previous and next

#include<iostream> using namespace std; void update(int arr[], int n) { //save current value of array i.e arr[0] at initial int previous = arr[0]; //updating the corrent value of array arr[0] = arr[0] * arr[1]; //updating remaining of the array element befor the last element for (int i = 1; i < n - 1; i++){ int current = arr[i]; //saving the current element for next iteretion //multiply previous and next element and save to current element arr[i] = previous * arr[i + 1]; previous = current; //updating previous value for next iteration } //for last element arr[n-1] = previous * arr[n-1]; } int main() { int arr[50], n; cout<<"Enter the size of array: "; cin>>n; if(n <= 1){ cout<<"Undefined size for this problem!"; exit(0); } cout<<"Input in the array: "; for(int i = 0; i < n; i++) cin>>arr[i]; update(arr, n); //printing the modified array cout<<"Modified array: "; for (int i=0; i<n; i++) cout << arr[i] << " "; return 0; }

**Output: **

Enter the size of array: 7 Input in the array: 2 5 8 6 7 3 4 Modified array: 10 16 30 56 18 28 12

**Time Complexity: **O(n), where **n** is a number of elements in the array.

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