Reemplace cada elemento de la array por la multiplicación de anterior y siguiente

Dada una array de enteros, actualice cada elemento con la multiplicación de los elementos anteriores y siguientes con las siguientes excepciones. 

• a) El primer elemento se reemplaza por la multiplicación del primero y el segundo. 
• b) El último elemento se sustituye por la multiplicación del último y el penúltimo.

Ejemplo:  

Input: arr[] = {2, 3, 4, 5, 6}
Output: arr[] = {6, 8, 15, 24, 30}

// We get the above output using following
// arr[] = {2*3, 2*4, 3*5, 4*6, 5*6} 

Source: Top 25 Interview Questions

Una solución simple es crear una array auxiliar, copiar el contenido de la array dada a la array auxiliar. Finalmente, recorra la array auxiliar y actualice la array dada utilizando el valor copiado. La complejidad timeTime de esta solución es O(n), pero requiere O(n) espacio adicional.
Una solución eficiente puede resolver el problema en O(n) tiempo y O(1) espacio. La idea es realizar un seguimiento de los elementos anteriores en el ciclo. 

Diagrama de flujo:

Diagrama de flujo- modificar

A continuación se muestra la implementación de esta idea. 

// C++ program to update every array element with
// multiplication of previous and next numbers in array
#include<iostream>
using namespace std;
 
void modify(int arr[], int n)
{
    // Nothing to do when array size is 1
    if (n <= 1)
      return;
 
    // store current value of arr[0] and update it
    int prev = arr[0];
    arr[0] = arr[0] * arr[1];
 
    // Update rest of the array elements
    for (int i=1; i<n-1; i++)
    {
        // Store current value of next interaction
        int curr = arr[i];
 
        // Update current value using previous value
        arr[i] = prev * arr[i+1];
 
        // Update previous value
        prev = curr;
    }
 
    // Update last array element
    arr[n-1] = prev * arr[n-1];
}
 
// Driver program
int main()
{
    int arr[] = {2, 3, 4, 5, 6};
    int n = sizeof(arr)/sizeof(arr[0]);
    modify(arr, n);
    for (int i=0; i<n; i++)
      cout << arr[i] << " ";
    return 0;
}

// Java program to update every array element with
// multiplication of previous and next numbers in array
import java.io.*;
import java.util.*;
import java.lang.Math;
 
class Multiply
{
   static void modify(int arr[], int n)
    {
        // Nothing to do when array size is 1
        if (n <= 1)
            return;
 
        // store current value of arr[0] and update it
        int prev = arr[0];
        arr[0] = arr[0] * arr[1];
 
        // Update rest of the array elements
        for (int i=1; i<n-1; i++)
        {
            // Store current value of next interaction
            int curr = arr[i];
 
            // Update current value using previous value
            arr[i] = prev * arr[i+1];
 
            // Update previous value
            prev = curr;
        }
 
        // Update last array element
        arr[n-1] = prev * arr[n-1];
    }
 
    // Driver program to test above function
    public static void main(String[] args)
    {
        int arr[] = {2, 3, 4, 5, 6};
        int n = arr.length;
        modify(arr, n);
        for (int i=0; i<n; i++)
         System.out.print(arr[i]+" ");
    }
}
/* This code is contributed by Devesh Agrawal */

# Python program to update every array element with
# multiplication of previous and next numbers in array
 
def modify(arr, n):
 
 # Nothing to do when array size is 1
    if n <= 1:
        return
 
    # store current value of arr[0] and update it
    prev = arr[0]
    arr[0] = arr[0] * arr[1]
 
    # Update rest of the array elements
    for i in range(1, n-1):
     
        # Store current value of next interaction
        curr = arr[i];
 
        # Update current value using previous value
        arr[i] = prev * arr[i+1]
 
           # Update previous value
        prev = curr
     
 
    # Update last array element
    arr[n-1] = prev * arr[n-1]
 
 
# Driver program
arr = [2, 3, 4, 5, 6]
n = len(arr)
modify(arr, n)
for i in range (0, n):
    print(arr[i],end=" ")
 
# This code is contributed by
# Smitha Dinesh Semwal

// C# program to update every array
// element with multiplication of
// previous and next numbers in array
using System;
 
class GFG
{
    static void modify(int []arr, int n)
    {
        // Nothing to do when array size is 1
        if (n <= 1)
            return;
 
        // store current value of arr[0] and update it
        int prev = arr[0];
        arr[0] = arr[0] * arr[1];
 
        // Update rest of the array elements
        for (int i=1; i<n-1; i++)
        {
            // Store current value of next interaction
            int curr = arr[i];
 
            // Update current value using previous value
            arr[i] = prev * arr[i+1];
 
            // Update previous value
            prev = curr;
        }
 
        // Update last array element
        arr[n-1] = prev * arr[n-1];
    }
 
    // Driver program to test above function
    public static void Main()
    {
        int []arr = {2, 3, 4, 5, 6};
        int n = arr.Length;
        modify(arr, n);
        for (int i=0; i<n; i++)
        Console.Write(arr[i]+" ");
    }
}
 
// This code is contributed by Sam007

<?php
// PHP program to update every array
// element with multiplication of previous
// and next numbers in array
 
function modify(&$arr, $n)
{
    // Nothing to do when array size is 1
    if ($n <= 1)
    return;
 
    // store current value of arr[0]
    // and update it
    $prev = $arr[0];
    $arr[0] = $arr[0] * $arr[1];
 
    // Update rest of the array elements
    for ($i = 1; $i < $n - 1; $i++)
    {
        // Store current value of
        // next interaction
        $curr = $arr[$i];
 
        // Update current value using
        // previous value
        $arr[$i] = $prev * $arr[$i + 1];
 
        // Update previous value
        $prev = $curr;
    }
 
    // Update last array element
    $arr[$n-1] = $prev * $arr[$n - 1];
}
 
// Driver Code
$arr = array (2, 3, 4, 5, 6);
$n = sizeof($arr);
modify($arr, $n);
for ($i = 0; $i < $n; $i++)
echo $arr[$i] ." ";
 
// This code is contributed
// by ChitraNayal
?>

<script>
// Javascript program to update every array element with
// multiplication of previous and next numbers in array
 
function modify(arr, n)
{
    // Nothing to do when array size is 1
    if (n <= 1)
      return;
 
    // store current value of arr[0] and update it
    let prev = arr[0];
    arr[0] = arr[0] * arr[1];
 
    // Update rest of the array elements
    for (let i = 1; i < n - 1; i++)
    {
        // Store current value of next interaction
        let curr = arr[i];
 
        // Update current value using previous value
        arr[i] = prev * arr[i+1];
 
        // Update previous value
        prev = curr;
    }
 
    // Update last array element
    arr[n-1] = prev * arr[n-1];
}
 
// Driver program
let arr = [2, 3, 4, 5, 6];
let n = arr.length;
modify(arr, n);
for (let i = 0; i < n; i++)
    document.write(arr[i] + " ");
     
// This code is contributed by subham348.
</script>

Producción: 

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Este artículo es una contribución de Ravi . Escriba comentarios si encuentra algo incorrecto, si desea compartir más información sobre el tema tratado anteriormente.