Dada una array de enteros positivos. Estamos obligados a escribir un programa para imprimir el producto mínimo de dos números cualesquiera de la array dada.
Ejemplos:
Input : 11 8 5 7 5 100 Output : 25 Explanation : The minimum product of any two numbers will be 5 * 5 = 25. Input : 198 76 544 123 154 675 Output : 7448 Explanation : The minimum product of any two numbers will be 76 * 123 = 7448.
Enfoque simple: un enfoque simple será ejecutar dos bucles anidados para generar todos los pares de elementos posibles y realizar un seguimiento del producto mínimo.
Complejidad de Tiempo: O( n * n)
Espacio Auxiliar: O( 1 )
Mejor enfoque: un enfoque eficiente será ordenar primero la array dada e imprimir el producto de los dos primeros números, la clasificación tomará O (n log n). La respuesta será entonces a[0] * a[1]
Complejidad de tiempo: O( n * log(n))
Espacio auxiliar: O( 1 )
Mejor enfoque: la idea es atravesar linealmente una array dada y realizar un seguimiento de un mínimo de dos elementos. Finalmente retorno producto de dos elementos mínimos.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ program to calculate minimum
// product of a pair
#include <bits/stdc++.h>
using namespace std;
// Function to calculate minimum product
// of pair
int printMinimumProduct(int arr[], int n)
{
// Initialize first and second
// minimums. It is assumed that the
// array has at least two elements.
int first_min = min(arr[0], arr[1]);
int second_min = max(arr[0], arr[1]);
// Traverse remaining array and keep
// track of two minimum elements (Note
// that the two minimum elements may
// be same if minimum element appears
// more than once)
// more than once)
for (int i=2; i<n; i++)
{
if (arr[i] < first_min)
{
second_min = first_min;
first_min = arr[i];
}
else if (arr[i] < second_min)
second_min = arr[i];
}
return first_min * second_min;
}
// Driver program to test above function
int main()
{
int a[] = { 11, 8 , 5 , 7 , 5 , 100 };
int n = sizeof(a) / sizeof(a[0]);
cout << printMinimumProduct(a,n);
return 0;
}
Java
// Java program to calculate minimum
// product of a pair
import java.util.*;
class GFG {
// Function to calculate minimum product
// of pair
static int printMinimumProduct(int arr[], int n)
{
// Initialize first and second
// minimums. It is assumed that the
// array has at least two elements.
int first_min = Math.min(arr[0], arr[1]);
int second_min = Math.max(arr[0], arr[1]);
// Traverse remaining array and keep
// track of two minimum elements (Note
// that the two minimum elements may
// be same if minimum element appears
// more than once)
// more than once)
for (int i = 2; i < n; i++)
{
if (arr[i] < first_min)
{
second_min = first_min;
first_min = arr[i];
}
else if (arr[i] < second_min)
second_min = arr[i];
}
return first_min * second_min;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int a[] = { 11, 8 , 5 , 7 , 5 , 100 };
int n = a.length;
System.out.print(printMinimumProduct(a,n));
}
}
// This code is contributed by Arnav Kr. Mandal.
Python3
# Python program to # calculate minimum # product of a pair # Function to calculate # minimum product # of pair def printMinimumProduct(arr,n): # Initialize first and second # minimums. It is assumed that the # array has at least two elements. first_min = min(arr[0], arr[1]) second_min = max(arr[0], arr[1]) # Traverse remaining array and keep # track of two minimum elements (Note # that the two minimum elements may # be same if minimum element appears # more than once) # more than once) for i in range(2,n): if (arr[i] < first_min): second_min = first_min first_min = arr[i] else if (arr[i] < second_min): second_min = arr[i] return first_min * second_min # Driver code a= [ 11, 8 , 5 , 7 , 5 , 100 ] n = len(a) print(printMinimumProduct(a,n)) # This code is contributed # by Anant Agarwal.
C#
// C# program to calculate minimum
// product of a pair
using System;
class GFG {
// Function to calculate minimum
// product of pair
static int printMinimumProduct(int []arr,
int n)
{
// Initialize first and second
// minimums. It is assumed that
// the array has at least two
// elements.
int first_min = Math.Min(arr[0],
arr[1]);
int second_min = Math.Max(arr[0],
arr[1]);
// Traverse remaining array and
// keep track of two minimum
// elements (Note that the two
// minimum elements may be same
// if minimum element appears
// more than once)
for (int i = 2; i < n; i++)
{
if (arr[i] < first_min)
{
second_min = first_min;
first_min = arr[i];
}
else if (arr[i] < second_min)
second_min = arr[i];
}
return first_min * second_min;
}
/* Driver program to test above
function */
public static void Main()
{
int []a = { 11, 8 , 5 , 7 ,
5 , 100 };
int n = a.Length;
Console.WriteLine(
printMinimumProduct(a, n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to calculate minimum
// product of a pair
// Function to calculate minimum
// product of pair
function printMinimumProduct($arr, $n)
{
// Initialize first and second
// minimums. It is assumed that the
// array has at least two elements.
$first_min = min($arr[0], $arr[1]);
$second_min = max($arr[0], $arr[1]);
// Traverse remaining array and keep
// track of two minimum elements (Note
// that the two minimum elements may
// be same if minimum element appears
// more than once)
// more than once)
for ($i = 2; $i < $n; $i++)
{
if ($arr[$i] < $first_min)
{
$second_min = $first_min;
$first_min = $arr[$i];
}
else if ($arr[$i] < $second_min)
$second_min = $arr[$i];
}
return $first_min * $second_min;
}
// Driver Code
$a = array(11, 8 , 5 , 7 , 5 , 100);
$n = sizeof($a);
echo(printMinimumProduct($a, $n));
// This code is contributed by Ajit.
?>
Javascript
<script>
// Javascript program to calculate minimum
// product of a pair
// Function to calculate minimum product
// of pair
function printMinimumProduct(arr, n)
{
// Initialize first and second
// minimums. It is assumed that the
// array has at least two elements.
let first_min = Math.min(arr[0], arr[1]);
let second_min = Math.max(arr[0], arr[1]);
// Traverse remaining array and keep
// track of two minimum elements (Note
// that the two minimum elements may
// be same if minimum element appears
// more than once)
// more than once)
for (let i=2; i<n; i++)
{
if (arr[i] < first_min)
{
second_min = first_min;
first_min = arr[i];
}
else if (arr[i] < second_min)
second_min = arr[i];
}
return first_min * second_min;
}
// Driver program to test above function
let a = [ 11, 8 , 5 , 7 , 5 , 100 ];
let n = a.length;
document.write(printMinimumProduct(a,n));
// This code is contributed by Mayank Tyagi
</script>
Producción
25
Complejidad temporal: O(n)
Espacio auxiliar: O(1)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA