Find a pair with maximum product in array
Here given code implementation process.
// C Program
// Find a pair with maximum product in array
#include <stdio.h>
#include <limits.h>
//Function which is display array elements
void display(int arr[], int size)
{
for (int i = 0; i < size; ++i)
{
printf(" %d", arr[i]);
}
printf("\n");
}
//Find maximum product of given unsorted array
void maximum_product(int arr[], int size)
{
if (size < 2)
{
return;
}
//Display array element
printf("\n Array :");
display(arr, size);
// Variable which is used to store two smallest element
int small_a = INT_MAX;
int small_b = INT_MAX;
// Variable which is used to store two largest element
int big_a = INT_MIN;
int big_b = INT_MIN;
for (int i = 0; i < size ; i++)
{
//Find two large numbers
if (big_a < arr[i])
{
if (big_b < big_a)
{
big_b = big_a;
}
big_a = arr[i];
}
else if (big_b < arr[i])
{
big_b = arr[i];
}
//Find two small numbers
//This calculations are used when number is negative
if (small_a > arr[i])
{
if (small_b > small_a)
{
small_b = small_a;
}
small_a = arr[i];
}
else if (small_b > arr[i])
{
small_b = arr[i];
}
}
if (small_a *small_b > big_a *big_b)
{
printf(" Maximum product [(%d)*(%d)] : %d\n", small_a, small_b, small_a *small_b);
}
else
{
printf(" Maximum product [(%d)*(%d)] : %d\n", big_a, big_b, (big_a *big_b));
}
}
int main()
{
//Define array of integer elements
int arr1[] = {
1 , 4 , 8 , 3 , 5 , 2
};
//Get the size of array
int size = sizeof(arr1) / sizeof(arr1[0]);
maximum_product(arr1, size);
int arr2[] = {
6 , 3 , -4 , -7 , 2 , 4
};
//Get the size of array
size = sizeof(arr2) / sizeof(arr2[0]);
maximum_product(arr2, size);
int arr3[] = {
1 , 8 , 2 , -4 , -3
};
//Get the size of array
size = sizeof(arr3) / sizeof(arr3[0]);
maximum_product(arr3, size);
return 0;
}Output
Array : 1 4 8 3 5 2
Maximum product [(8)*(5)] : 40
Array : 6 3 -4 -7 2 4
Maximum product [(-7)*(-4)] : 28
Array : 1 8 2 -4 -3
Maximum product [(8)*(2)] : 16/*
Java Program
Find a pair with maximum product in array
*/
class MyArray
{
//Function which is display array elements
public void display(int[] arr, int size)
{
for (int i = 0; i < size; ++i)
{
System.out.print(" " + arr[i]);
}
System.out.print("\n");
}
//Find maximum product of given unsorted array
public void maximum_product(int[] arr, int size)
{
if (size < 2)
{
return;
}
System.out.print("\n Array :");
display(arr, size);
// Variable which is used to store two smallest element
int small_a = Integer.MAX_VALUE;
int small_b = Integer.MAX_VALUE;
// Variable which is used to store two largest element
int big_a = Integer.MIN_VALUE;
int big_b = Integer.MIN_VALUE;
for (int i = 0; i < size; i++)
{
//Find two large numbers
if (big_a < arr[i])
{
if (big_b < big_a)
{
big_b = big_a;
}
big_a = arr[i];
}
else if (big_b < arr[i])
{
big_b = arr[i];
}
//Find two small numbers
//This calculations are used when number is negative
if (small_a > arr[i])
{
if (small_b > small_a)
{
small_b = small_a;
}
small_a = arr[i];
}
else if (small_b > arr[i])
{
small_b = arr[i];
}
}
if (small_a * small_b > big_a * big_b)
{
System.out.print(" Maximum product [(" + small_a + ")*(" + small_b + ")] : " + small_a * small_b + "\n");
}
else
{
System.out.print(" Maximum product [(" + big_a + ")*(" + big_b + ")] : " + (big_a * big_b) + "\n");
}
}
public static void main(String[] args)
{
MyArray obj = new MyArray();
//Define array of integer elements
int[] arr1 = {
1,
4,
8,
3,
5,
2
};
//Get the size of array
int size = arr1.length;
obj.maximum_product(arr1, size);
int[] arr2 = {
6,
3,
-4,
-7,
2,
4
};
//Get the size of array
size = arr2.length;
obj.maximum_product(arr2, size);
int[] arr3 = {
1,
8,
2,
-4,
-3
};
//Get the size of array
size = arr3.length;
obj.maximum_product(arr3, size);
}
}Output
Array : 1 4 8 3 5 2
Maximum product [(8)*(5)] : 40
Array : 6 3 -4 -7 2 4
Maximum product [(-7)*(-4)] : 28
Array : 1 8 2 -4 -3
Maximum product [(8)*(2)] : 16//Include header file
#include <iostream>
#include<limits.h>
using namespace std;
/*
C++ Program
Find a pair with maximum product in array
*/
class MyArray
{
public:
//Function which is display array elements
void display(int arr[], int size)
{
for (int i = 0; i < size; ++i)
{
cout << " " << arr[i];
}
cout << "\n";
}
//Find maximum product of given unsorted array
void maximum_product(int arr[], int size)
{
if (size < 2)
{
return;
}
cout << "\n Array :";
this->display(arr, size);
// Variable which is used to store two smallest element
int small_a = INT_MAX;
int small_b = INT_MAX;
// Variable which is used to store two largest element
int big_a = INT_MIN;
int big_b = INT_MIN;
for (int i = 0; i < size; i++)
{
//Find two large numbers
if (big_a < arr[i])
{
if (big_b < big_a)
{
big_b = big_a;
}
big_a = arr[i];
}
else if (big_b < arr[i])
{
big_b = arr[i];
}
//Find two small numbers
//This calculations are used when number is negative
if (small_a > arr[i])
{
if (small_b > small_a)
{
small_b = small_a;
}
small_a = arr[i];
}
else if (small_b > arr[i])
{
small_b = arr[i];
}
}
if (small_a * small_b > big_a * big_b)
{
cout << " Maximum product [(" << small_a << ")*(" << small_b << ")] : " << small_a * small_b << "\n";
}
else
{
cout << " Maximum product [(" << big_a << ")*(" << big_b << ")] : " << (big_a * big_b) << "\n";
}
}
};
int main()
{
MyArray obj = MyArray();
int arr1[] = {
1 , 4 , 8 , 3 , 5 , 2
};
//Get the size of array
int size = sizeof(arr1) / sizeof(arr1[0]);
obj.maximum_product(arr1, size);
int arr2[] = {
6 , 3 , -4 , -7 , 2 , 4
};
//Get the size of array
size = sizeof(arr2) / sizeof(arr2[0]);
obj.maximum_product(arr2, size);
int arr3[] = {
1 , 8 , 2 , -4 , -3
};
//Get the size of array
size = sizeof(arr3) / sizeof(arr3[0]);
obj.maximum_product(arr3, size);
return 0;
}Output
Array : 1 4 8 3 5 2
Maximum product [(8)*(5)] : 40
Array : 6 3 -4 -7 2 4
Maximum product [(-7)*(-4)] : 28
Array : 1 8 2 -4 -3
Maximum product [(8)*(2)] : 16//Include namespace system
using System;
/*
C# Program
Find a pair with maximum product in array
*/
class MyArray
{
//Function which is display array elements
public void display(int[] arr, int size)
{
for (int i = 0; i < size; ++i)
{
Console.Write(" " + arr[i]);
}
Console.Write("\n");
}
//Find maximum product of given unsorted array
public void maximum_product(int[] arr, int size)
{
if (size < 2)
{
return;
}
Console.Write("\n Array :");
display(arr, size);
// Variable which is used to store two smallest element
int small_a = int.MaxValue;
int small_b = int.MaxValue;
// Variable which is used to store two largest element
int big_a = int.MinValue;
int big_b = int.MinValue;
for (int i = 0; i < size; i++)
{
//Find two large numbers
if (big_a < arr[i])
{
if (big_b < big_a)
{
big_b = big_a;
}
big_a = arr[i];
}
else if (big_b < arr[i])
{
big_b = arr[i];
}
//Find two small numbers
//This calculations are used when number is negative
if (small_a > arr[i])
{
if (small_b > small_a)
{
small_b = small_a;
}
small_a = arr[i];
}
else if (small_b > arr[i])
{
small_b = arr[i];
}
}
if (small_a * small_b > big_a * big_b)
{
Console.Write(" Maximum product [(" + small_a + ")*(" + small_b + ")] : " + small_a * small_b + "\n");
}
else
{
Console.Write(" Maximum product [(" + big_a + ")*(" + big_b + ")] : " + (big_a * big_b) + "\n");
}
}
public static void Main(String[] args)
{
MyArray obj = new MyArray();
int[] arr1 = {
1 , 4 , 8 , 3 , 5 , 2
};
//Get the size of array
int size = arr1.Length;
obj.maximum_product(arr1, size);
int[] arr2 = {
6 , 3 , -4 , -7 , 2 , 4
};
//Get the size of array
size = arr2.Length;
obj.maximum_product(arr2, size);
int[] arr3 = {
1 , 8 , 2 , -4 , -3
};
//Get the size of array
size = arr3.Length;
obj.maximum_product(arr3, size);
}
}Output
Array : 1 4 8 3 5 2
Maximum product [(8)*(5)] : 40
Array : 6 3 -4 -7 2 4
Maximum product [(-7)*(-4)] : 28
Array : 1 8 2 -4 -3
Maximum product [(8)*(2)] : 16<?php
/*
Php Program
Find a pair with maximum product in array
*/
class MyArray
{
//Function which is display array elements
public function display( $arr, $size)
{
for ($i = 0; $i < $size; ++$i)
{
echo " ". $arr[$i];
}
echo "\n";
}
//Find maximum product of given unsorted array
public function maximum_product( $arr, $size)
{
if ($size < 2)
{
return;
}
echo "\n Array :";
$this->display($arr, $size);
// Variable which is used to store two smallest element
$small_a = PHP_INT_MAX;
$small_b = PHP_INT_MAX;
// Variable which is used to store two largest element
$big_a = -PHP_INT_MAX;
$big_b = -PHP_INT_MAX;
for ($i = 0; $i < $size; $i++)
{
//Find two large numbers
if ($big_a < $arr[$i])
{
if ($big_b < $big_a)
{
$big_b = $big_a;
}
$big_a = $arr[$i];
}
else if ($big_b < $arr[$i])
{
$big_b = $arr[$i];
}
//Find two small numbers
//This calculations are used when number is negative
if ($small_a > $arr[$i])
{
if ($small_b > $small_a)
{
$small_b = $small_a;
}
$small_a = $arr[$i];
}
else if ($small_b > $arr[$i])
{
$small_b = $arr[$i];
}
}
if ($small_a * $small_b > $big_a * $big_b)
{
echo " Maximum product [(". $small_a .")*(". $small_b .")] : ". $small_a * $small_b ."\n";
}
else
{
echo " Maximum product [(". $big_a .")*(". $big_b .")] : ". ($big_a * $big_b) ."\n";
}
}
}
function main()
{
$obj = new MyArray();
//Define array of integer elements
$arr1 = array(1, 4, 8, 3, 5, 2);
//Get the size of array
$size = count($arr1);
$obj->maximum_product($arr1, $size);
$arr2 = array(6, 3, -4, -7, 2, 4);
//Get the size of array
$size = count($arr2);
$obj->maximum_product($arr2, $size);
$arr3 = array(1, 8, 2, -4, -3);
//Get the size of array
$size = count($arr3);
$obj->maximum_product($arr3, $size);
}
main();Output
Array : 1 4 8 3 5 2
Maximum product [(8)*(5)] : 40
Array : 6 3 -4 -7 2 4
Maximum product [(-7)*(-4)] : 28
Array : 1 8 2 -4 -3
Maximum product [(8)*(2)] : 16/*
Node Js Program
Find a pair with maximum product in array
*/
class MyArray
{
//Function which is display array elements
display(arr, size)
{
for (var i = 0; i < size; ++i)
{
process.stdout.write(" " + arr[i]);
}
process.stdout.write("\n");
}
//Find maximum product of given unsorted array
maximum_product(arr, size)
{
if (size < 2)
{
return;
}
process.stdout.write("\n Array :");
this.display(arr, size);
// Variable which is used to store two smallest element
var small_a = Number.MAX_VALUE;
var small_b = Number.MAX_VALUE;
// Variable which is used to store two largest element
var big_a = -Number.MAX_VALUE;
var big_b = -Number.MAX_VALUE;
for (var i = 0; i < size; i++)
{
//Find two large numbers
if (big_a < arr[i])
{
if (big_b < big_a)
{
big_b = big_a;
}
big_a = arr[i];
}
else if (big_b < arr[i])
{
big_b = arr[i];
}
//Find two small numbers
//This calculations are used when number is negative
if (small_a > arr[i])
{
if (small_b > small_a)
{
small_b = small_a;
}
small_a = arr[i];
}
else if (small_b > arr[i])
{
small_b = arr[i];
}
}
if (small_a * small_b > big_a * big_b)
{
process.stdout.write(" Maximum product [(" + small_a + ")*(" + small_b + ")] : " + small_a * small_b + "\n");
}
else
{
process.stdout.write(" Maximum product [(" + big_a + ")*(" + big_b + ")] : " + (big_a * big_b) + "\n");
}
}
}
function main()
{
var obj = new MyArray();
//Define array of integer elements
var arr1 = [1, 4, 8, 3, 5, 2];
//Get the size of array
var size = arr1.length;
obj.maximum_product(arr1, size);
var arr2 = [6, 3, -4, -7, 2, 4];
//Get the size of array
size = arr2.length;
obj.maximum_product(arr2, size);
var arr3 = [1, 8, 2, -4, -3];
//Get the size of array
size = arr3.length;
obj.maximum_product(arr3, size);
}
main();Output
Array : 1 4 8 3 5 2
Maximum product [(8)*(5)] : 40
Array : 6 3 -4 -7 2 4
Maximum product [(-7)*(-4)] : 28
Array : 1 8 2 -4 -3
Maximum product [(8)*(2)] : 16import sys
# Python 3 Program
# Find a pair with maximum product in array
class MyArray :
# Function which is display array elements
def display(self, arr, size) :
i = 0
while (i < size) :
print(" ", arr[i], end = "")
i += 1
print("\n", end = "")
# Find maximum product of given unsorted array
def maximum_product(self, arr, size) :
if (size < 2) :
return
print("\n Array :", end = "")
self.display(arr, size)
# Variable which is used to store two smallest element
small_a = sys.maxsize
small_b = sys.maxsize
# Variable which is used to store two largest element
big_a = -sys.maxsize
big_b = -sys.maxsize
i = 0
while (i < size) :
# Find two large numbers
if (big_a < arr[i]) :
if (big_b < big_a) :
big_b = big_a
big_a = arr[i]
elif(big_b < arr[i]) :
big_b = arr[i]
# Find two small numbers
# This calculations are used when number is negative
if (small_a > arr[i]) :
if (small_b > small_a) :
small_b = small_a
small_a = arr[i]
elif(small_b > arr[i]) :
small_b = arr[i]
i += 1
if (small_a * small_b > big_a * big_b) :
print(" Maximum product [(", small_a ,")*(", small_b ,")] : ", small_a * small_b ,"\n", end = "")
else :
print(" Maximum product [(", big_a ,")*(", big_b ,")] : ", (big_a * big_b) ,"\n", end = "")
def main() :
obj = MyArray()
# Define array of integer elements
arr1 = [1, 4, 8, 3, 5, 2]
# Get the size of array
size = len(arr1)
obj.maximum_product(arr1, size)
arr2 = [6, 3, -4, -7, 2, 4]
# Get the size of array
size = len(arr2)
obj.maximum_product(arr2, size)
arr3 = [1, 8, 2, -4, -3]
# Get the size of array
size = len(arr3)
obj.maximum_product(arr3, size)
if __name__ == "__main__": main()Output
Array : 1 4 8 3 5 2
Maximum product [( 8 )*( 5 )] : 40
Array : 6 3 -4 -7 2 4
Maximum product [( -7 )*( -4 )] : 28
Array : 1 8 2 -4 -3
Maximum product [( 8 )*( 2 )] : 16# Ruby Program
# Find a pair with maximum product in array
class MyArray
# Function which is display array elements
def display(arr, size)
i = 0
while (i < size)
print(" ", arr[i])
i += 1
end
print("\n")
end
# Find maximum product of given unsorted array
def maximum_product(arr, size)
if (size < 2)
return
end
print("\n Array :")
self.display(arr, size)
# Variable which is used to store two smallest element
small_a = (2 ** (0. size * 8 - 2))
small_b = (2 ** (0. size * 8 - 2))
# Variable which is used to store two largest element
big_a = -(2 ** (0. size * 8 - 2))
big_b = -(2 ** (0. size * 8 - 2))
i = 0
while (i < size)
# Find two large numbers
if (big_a < arr[i])
if (big_b < big_a)
big_b = big_a
end
big_a = arr[i]
elsif(big_b < arr[i])
big_b = arr[i]
end
# Find two small numbers
# This calculations are used when number is negative
if (small_a > arr[i])
if (small_b > small_a)
small_b = small_a
end
small_a = arr[i]
elsif(small_b > arr[i])
small_b = arr[i]
end
i += 1
end
if (small_a * small_b > big_a * big_b)
print(" Maximum product [(", small_a ,")*(", small_b ,")] : ", small_a * small_b ,"\n")
else
print(" Maximum product [(", big_a ,")*(", big_b ,")] : ", (big_a * big_b) ,"\n")
end
end
end
def main()
obj = MyArray.new()
# Define array of integer elements
arr1 = [1, 4, 8, 3, 5, 2]
# Get the size of array
size = arr1.length
obj.maximum_product(arr1, size)
arr2 = [6, 3, -4, -7, 2, 4]
# Get the size of array
size = arr2.length
obj.maximum_product(arr2, size)
arr3 = [1, 8, 2, -4, -3]
# Get the size of array
size = arr3.length
obj.maximum_product(arr3, size)
end
main()Output
Array : 1 4 8 3 5 2
Maximum product [(8)*(5)] : 40
Array : 6 3 -4 -7 2 4
Maximum product [(-7)*(-4)] : 28
Array : 1 8 2 -4 -3
Maximum product [(8)*(2)] : 16
/*
Scala Program
Find a pair with maximum product in array
*/
class MyArray
{
//Function which is display array elements
def display(arr: Array[Int], size: Int): Unit = {
var i: Int = 0;
while (i < size)
{
print(" " + arr(i));
i += 1;
}
print("\n");
}
//Find maximum product of given unsorted array
def maximum_product(arr: Array[Int], size: Int): Unit = {
if (size < 2)
{
return;
}
print("\n Array :");
display(arr, size);
// Variable which is used to store two smallest element
var small_a: Int = Int.MaxValue;
var small_b: Int = Int.MaxValue;
// Variable which is used to store two largest element
var big_a: Int = Int.MinValue;
var big_b: Int = Int.MinValue;
var i: Int = 0;
while (i < size)
{
//Find two large numbers
if (big_a < arr(i))
{
if (big_b < big_a)
{
big_b = big_a;
}
big_a = arr(i);
}
else if (big_b < arr(i))
{
big_b = arr(i);
}
//Find two small numbers
//This calculations are used when number is negative
if (small_a > arr(i))
{
if (small_b > small_a)
{
small_b = small_a;
}
small_a = arr(i);
}
else if (small_b > arr(i))
{
small_b = arr(i);
}
i += 1;
}
if (small_a * small_b > big_a * big_b)
{
print(" Maximum product [(" + small_a + ")*(" + small_b + ")] : " + small_a * small_b + "\n");
}
else
{
print(" Maximum product [(" + big_a + ")*(" + big_b + ")] : " + (big_a * big_b) + "\n");
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
var obj: MyArray = new MyArray();
//Define array of integer elements
var arr1: Array[Int] = Array(1, 4, 8, 3, 5, 2);
//Get the size of array
var size: Int = arr1.length;
obj.maximum_product(arr1, size);
var arr2: Array[Int] = Array(6, 3, -4, -7, 2, 4);
//Get the size of array
size = arr2.length;
obj.maximum_product(arr2, size);
var arr3: Array[Int] = Array(1, 8, 2, -4, -3);
//Get the size of array
size = arr3.length;
obj.maximum_product(arr3, size);
}
}Output
Array : 1 4 8 3 5 2
Maximum product [(8)*(5)] : 40
Array : 6 3 -4 -7 2 4
Maximum product [(-7)*(-4)] : 28
Array : 1 8 2 -4 -3
Maximum product [(8)*(2)] : 16/*
Swift Program
Find a pair with maximum product in array
*/
class MyArray
{
//Function which is display array elements
func display(_ arr: [Int], _ size: Int)
{
var i: Int = 0;
while (i < size)
{
print(" ", arr[i], terminator: "");
i += 1;
}
print("\n", terminator: "");
}
//Find maximum product of given unsorted array
func maximum_product(_ arr: [Int], _ size: Int)
{
if (size < 2)
{
return;
}
print("\n Array :", terminator: "");
self.display(arr, size);
// Variable which is used to store two smallest element
var small_a: Int = Int.max;
var small_b: Int = Int.max;
// Variable which is used to store two largest element
var big_a: Int = Int.min;
var big_b: Int = Int.min;
var i: Int = 0;
while (i < size)
{
//Find two large numbers
if (big_a < arr[i])
{
if (big_b < big_a)
{
big_b = big_a;
}
big_a = arr[i];
}
else if (big_b < arr[i])
{
big_b = arr[i];
}
//Find two small numbers
//This calculations are used when number is negative
if (small_a > arr[i])
{
if (small_b > small_a)
{
small_b = small_a;
}
small_a = arr[i];
}
else if (small_b > arr[i])
{
small_b = arr[i];
}
i += 1;
}
if (small_a * small_b > big_a * big_b)
{
print(" Maximum product [(", small_a ,")*(", small_b ,")] : ", small_a * small_b ,"\n", terminator: "");
}
else
{
print(" Maximum product [(", big_a ,")*(", big_b ,")] : ", (big_a * big_b) ,"\n", terminator: "");
}
}
}
func main()
{
let obj: MyArray = MyArray();
//Define array of integer elements
let arr1: [Int] = [1, 4, 8, 3, 5, 2];
//Get the size of array
var size: Int = arr1.count;
obj.maximum_product(arr1, size);
let arr2: [Int] = [6, 3, -4, -7, 2, 4];
//Get the size of array
size = arr2.count;
obj.maximum_product(arr2, size);
let arr3: [Int] = [1, 8, 2, -4, -3];
//Get the size of array
size = arr3.count;
obj.maximum_product(arr3, size);
}
main();Output
Array : 1 4 8 3 5 2
Maximum product [( 8 )*( 5 )] : 40
Array : 6 3 -4 -7 2 4
Maximum product [( -7 )*( -4 )] : 28
Array : 1 8 2 -4 -3
Maximum product [( 8 )*( 2 )] : 16
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