Dada una array arr[] de N enteros, la tarea es imprimir el elemento más grande entre la array de modo que el producto de su elemento anterior y siguiente sea máximo.
Ejemplos:
Entrada: arr[] = {5, 6, 4, 3, 2}
Salida: 6
El producto de los elementos siguiente y anterior
para cada elemento de la array dada es:
5 -> 2 * 6 = 12
6 -> 5 * 4 = 20
4 -> 6 * 3 = 18
3 -> 4 * 2 = 8
2 -> 3 * 5 = 15
De estos 20 es el máximo.
Por lo tanto, 6 es la respuesta.
Entrada: arr[] = {9, 2, 3, 1, 5, 17}
Salida: 17
Enfoque: para cada elemento de la array, encuentre el producto de su elemento anterior y siguiente. El elemento que tiene el producto máximo es el resultado. Si dos elementos tienen el mismo producto de los elementos siguientes y anteriores, elija el elemento mayor entre ellos.
A continuación se muestra la implementación del enfoque anterior:
C++
#include<bits/stdc++.h>
using namespace std;
// Function to return the largest element
// such that its previous and next
// element product is maximum
int maxElement(int a[], int n)
{
if (n < 3)
return -1;
int maxElement = a[0];
int maxProd = a[n - 1] * a[1];
for (int i = 1; i < n; i++)
{
// Calculate the product of the previous
// and the next element for
// the current element
int currProd = a[i - 1] * a[(i + 1) % n];
// Update the maximum product
if (currProd > maxProd)
{
maxProd = currProd;
maxElement = a[i];
}
// If current product is equal to the
// current maximum product then
// choose the maximum element
else if (currProd == maxProd)
{
maxElement = max(maxElement, a[i]);
}
}
return maxElement;
}
// Driver code
int main()
{
int a[] = { 5, 6, 4, 3, 2};
int n = sizeof(a)/sizeof(a[0]);
cout << maxElement(a, n);
return 0;
}
Java
// Java implementation of the approach
class GFG {
// Function to return the largest element
// such that its previous and next
// element product is maximum
static int maxElement(int a[], int n)
{
if (n < 3)
return -1;
int maxElement = a[0];
int maxProd = a[n - 1] * a[1];
for (int i = 1; i < n; i++) {
// Calculate the product of the previous
// and the next element for
// the current element
int currProd = a[i - 1] * a[(i + 1) % n];
// Update the maximum product
if (currProd > maxProd) {
maxProd = currProd;
maxElement = a[i];
}
// If current product is equal to the
// current maximum product then
// choose the maximum element
else if (currProd == maxProd) {
maxElement = Math.max(maxElement, a[i]);
}
}
return maxElement;
}
// Driver code
public static void main(String[] args)
{
int[] a = { 5, 6, 4, 3, 2 };
int n = a.length;
System.out.println(maxElement(a, n));
}
}
Python3
# Function to return the largest element # such that its previous and next # element product is maximum def maxElement(a, n): if n < 3: return -1 maxElement = a[0] maxProd = a[n - 1] * a[1] for i in range(1, n): # Calculate the product of the previous # and the next element for # the current element currprod = a[i - 1] * a[(i + 1) % n] if currprod > maxProd: maxProd = currprod maxElement = a[i] # If current product is equal to the # current maximum product then # choose the maximum element elif currprod == maxProd: maxElement = max(maxElement, a[i]) return maxElement # Driver code a = [5, 6, 4, 3, 2] n = len(a)#sizeof(a[0]) print(maxElement(a, n)) # This code is contributed by mohit kumar 29
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the largest element
// such that its previous and next
// element product is maximum
static int maxElement(int []a, int n)
{
if (n < 3)
return -1;
int maxElement = a[0];
int maxProd = a[n - 1] * a[1];
for (int i = 1; i < n; i++)
{
// Calculate the product of the previous
// and the next element for
// the current element
int currProd = a[i - 1] * a[(i + 1) % n];
// Update the maximum product
if (currProd > maxProd)
{
maxProd = currProd;
maxElement = a[i];
}
// If current product is equal to the
// current maximum product then
// choose the maximum element
else if (currProd == maxProd)
{
maxElement = Math.Max(maxElement, a[i]);
}
}
return maxElement;
}
// Driver code
public static void Main()
{
int[] a = { 5, 6, 4, 3, 2 };
int n = a.Length;
Console.WriteLine(maxElement(a, n));
}
}
// This code is contributed by AnkitRai01
Javascript
<script>
// Java script implementation of the approach
// Function to return the largest element
// such that its previous and next
// element product is maximum
function maxElement(a,n)
{
if (n < 3)
return -1;
let maxElement = a[0];
let maxProd = a[n - 1] * a[1];
for (let i = 1; i < n; i++) {
// Calculate the product of the previous
// and the next element for
// the current element
let currProd = a[i - 1] * a[(i + 1) % n];
// Update the maximum product
if (currProd > maxProd) {
maxProd = currProd;
maxElement = a[i];
}
// If current product is equal to the
// current maximum product then
// choose the maximum element
else if (currProd == maxProd) {
maxElement = Math.max(maxElement, a[i]);
}
}
return maxElement;
}
// Driver code
let a = [ 5, 6, 4, 3, 2 ];
let n = a.length;
document.write(maxElement(a, n));
// This code is contributed by sravan kumar G
</script>
Producción:
6
Complejidad temporal: O(n), ya que se ejecuta un ciclo por una vez de 1 a (n – 1).
Espacio Auxiliar: O(1), ya que no se ha tomado ningún espacio extra.