Posted on by Kalkicode
Code Mathematics

Two elements whose sum is closest to zero

"Two elements whose sum is closest to zero in an array" means finding any two numbers in the given array whose sum is closest to zero. In other words, the task is to find two numbers in the array whose sum is closest to zero, regardless of whether it is a positive or negative number. For example, if the given array is [4, 2, -1, 5, -7, 1], then the two elements whose sum is closest to zero are -1 and 1 because their sum is zero and they are the closest to zero compared to any other pair of numbers in the array.

/*
    C program for
    Two elements whose sum is closest to zero
*/
#include <stdio.h>

int absValue(int num)
{
	if (num < 0)
	{
		return -num;
	}
	return num;
}
void closestZeroSum(int arr[], int n)
{
	if (n <= 1)
	{
		return;
	}
	// Get value of first element
	int x = arr[0];
	// Get value of second element
	int y = arr[1];
	int sum = x + y;
	// Execute the loop through by size of array
	for (int i = 0; i < n; ++i)
	{
		for (int j = i + 1; j < n; ++j)
		{
			if (absValue(sum) > absValue(arr[i] + arr[j]))
			{
				// Get pair of closest to zero
				x = arr[i];
				y = arr[j];
				// Change sum
				sum = x + y;
			}
		}
	}
	// Display resultant pair
	printf("\n (%d) + (%d) = %d", x, y, sum);
}
int main()
{
	int arr1[] = {
		4 , 7 , 8 , 2 , -6 , 12
	};
	int arr2[] = {
		2 , 6 , 8 , -2 , 5
	};
	// Test A
	int n = sizeof(arr1) / sizeof(arr1[0]);
	closestZeroSum(arr1, n);
	// Test B
	n = sizeof(arr2) / sizeof(arr2[0]);
	closestZeroSum(arr2, n);
	return 0;
}

Output

 (7) + (-6) = 1
 (2) + (-2) = 0
/*
    Java program for
    Two elements whose sum is closest to zero
*/
public class Test
{
	public int absValue(int num)
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	public void closestZeroSum(int[] arr, int n)
	{
		if (n <= 1)
		{
			return;
		}
		// Get value of first element
		int x = arr[0];
		// Get value of second element
		int y = arr[1];
		int sum = x + y;
		// Execute the loop through by size of array
		for (int i = 0; i < n; ++i)
		{
			for (int j = i + 1; j < n; ++j)
			{
				if (absValue(sum) > absValue(arr[i] + arr[j]))
				{
					// Get pair of closest to zero
					x = arr[i];
					y = arr[j];
					// Change sum
					sum = x + y;
				}
			}
		}
		// Display resultant pair
		System.out.println(" (" + x + ") + (" + y + ") = " + sum);
	}
	public static void main(String[] args)
	{
		Test task = new Test();
		int[] arr1 = {
			4 , 7 , 8 , 2 , -6 , 12
		};
		int[] arr2 = {
			2 , 6 , 8 , -2 , 5
		};
		// Test A
		int n = arr1.length;
		task.closestZeroSum(arr1, n);
		// Test B
		n = arr2.length;
		task.closestZeroSum(arr2, n);
	}
}

Output

 (7) + (-6) = 1
 (2) + (-2) = 0
// Include header file
#include <iostream>
using namespace std;
/*
    C++ program for
    Two elements whose sum is closest to zero
*/
class Test
{
	public: int absValue(int num)
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	void closestZeroSum(int arr[], int n)
	{
		if (n <= 1)
		{
			return;
		}
		// Get value of first element
		int x = arr[0];
		// Get value of second element
		int y = arr[1];
		int sum = x + y;
		// Execute the loop through by size of array
		for (int i = 0; i < n; ++i)
		{
			for (int j = i + 1; j < n; ++j)
			{
				if (this->absValue(sum) > this->absValue(arr[i] + arr[j]))
				{
					// Get pair of closest to zero
					x = arr[i];
					y = arr[j];
					// Change sum
					sum = x + y;
				}
			}
		}
		// Display resultant pair
		cout << " (" << x << ") + (" << y << ") = " << sum << endl;
	}
};
int main()
{
	Test *task = new Test();
	int arr1[] = {
		4 , 7 , 8 , 2 , -6 , 12
	};
	int arr2[] = {
		2 , 6 , 8 , -2 , 5
	};
	// Test A
	int n = sizeof(arr1) / sizeof(arr1[0]);
	task->closestZeroSum(arr1, n);
	// Test B
	n = sizeof(arr2) / sizeof(arr2[0]);
	task->closestZeroSum(arr2, n);
	return 0;
}

Output

 (7) + (-6) = 1
 (2) + (-2) = 0
// Include namespace system
using System;
/*
    Csharp program for
    Two elements whose sum is closest to zero
*/
public class Test
{
	public int absValue(int num)
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	public void closestZeroSum(int[] arr, int n)
	{
		if (n <= 1)
		{
			return;
		}
		// Get value of first element
		int x = arr[0];
		// Get value of second element
		int y = arr[1];
		int sum = x + y;
		// Execute the loop through by size of array
		for (int i = 0; i < n; ++i)
		{
			for (int j = i + 1; j < n; ++j)
			{
				if (this.absValue(sum) > this.absValue(arr[i] + arr[j]))
				{
					// Get pair of closest to zero
					x = arr[i];
					y = arr[j];
					// Change sum
					sum = x + y;
				}
			}
		}
		// Display resultant pair
		Console.WriteLine(" (" + x + ") + (" + y + ") = " + sum);
	}
	public static void Main(String[] args)
	{
		Test task = new Test();
		int[] arr1 = {
			4 , 7 , 8 , 2 , -6 , 12
		};
		int[] arr2 = {
			2 , 6 , 8 , -2 , 5
		};
		// Test A
		int n = arr1.Length;
		task.closestZeroSum(arr1, n);
		// Test B
		n = arr2.Length;
		task.closestZeroSum(arr2, n);
	}
}

Output

 (7) + (-6) = 1
 (2) + (-2) = 0
package main
import "fmt"
/*
    Go program for
    Two elements whose sum is closest to zero
*/

func absValue(num int) int {
	if num < 0 {
		return -num
	}
	return num
}
func closestZeroSum(arr[] int, n int) {
	if n <= 1 {
		return
	}
	// Get value of first element
	var x int = arr[0]
	// Get value of second element
	var y int = arr[1]
	var sum int = x + y
	// Execute the loop through by size of array
	for i := 0 ; i < n ; i++ {
		for j := i + 1 ; j < n ; j++ {
			if absValue(sum) > absValue(arr[i] + arr[j]) {
				// Get pair of closest to zero
				x = arr[i]
				y = arr[j]
				// Change sum
				sum = x + y
			}
		}
	}
	// Display resultant pair
	fmt.Println(" (", x, ") + (", y, ") = ", sum)
}
func main() {

	var arr1 = [] int {4, 7, 8, 2, -6, 12}
	var arr2 = [] int {2, 6, 8, -2, 5}
	// Test A
	var n int = len(arr1)
	closestZeroSum(arr1, n)
	// Test B
	n = len(arr2)
	closestZeroSum(arr2, n)
}

Output

 (7) + (-6) = 1
 (2) + (-2) = 0
<?php
/*
    Php program for
    Two elements whose sum is closest to zero
*/
class Test
{
	public	function absValue($num)
	{
		if ($num < 0)
		{
			return -$num;
		}
		return $num;
	}
	public	function closestZeroSum($arr, $n)
	{
		if ($n <= 1)
		{
			return;
		}
		// Get value of first element
		$x = $arr[0];
		// Get value of second element
		$y = $arr[1];
		$sum = $x + $y;
		// Execute the loop through by size of array
		for ($i = 0; $i < $n; ++$i)
		{
			for ($j = $i + 1; $j < $n; ++$j)
			{
				if ($this->absValue($sum) > 
                    $this->absValue($arr[$i] + $arr[$j]))
				{
					// Get pair of closest to zero
					$x = $arr[$i];
					$y = $arr[$j];
					// Change sum
					$sum = $x + $y;
				}
			}
		}
		// Display resultant pair
		echo(" (".$x.
			") + (".$y.
			") = ".$sum.
			"\n");
	}
}

function main()
{
	$task = new Test();
	$arr1 = array(4, 7, 8, 2, -6, 12);
	$arr2 = array(2, 6, 8, -2, 5);
	// Test A
	$n = count($arr1);
	$task->closestZeroSum($arr1, $n);
	// Test B
	$n = count($arr2);
	$task->closestZeroSum($arr2, $n);
}
main();

Output

 (7) + (-6) = 1
 (2) + (-2) = 0
/*
    Node JS program for
    Two elements whose sum is closest to zero
*/
class Test
{
	absValue(num)
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	closestZeroSum(arr, n)
	{
		if (n <= 1)
		{
			return;
		}
		// Get value of first element
		var x = arr[0];
		// Get value of second element
		var y = arr[1];
		var sum = x + y;
		// Execute the loop through by size of array
		for (var i = 0; i < n; ++i)
		{
			for (var j = i + 1; j < n; ++j)
			{
				if (this.absValue(sum) > this.absValue(arr[i] + arr[j]))
				{
					// Get pair of closest to zero
					x = arr[i];
					y = arr[j];
					// Change sum
					sum = x + y;
				}
			}
		}
		// Display resultant pair
		console.log(" (" + x + ") + (" + y + ") = " + sum);
	}
}

function main()
{
	var task = new Test();
	var arr1 = [4, 7, 8, 2, -6, 12];
	var arr2 = [2, 6, 8, -2, 5];
	// Test A
	var n = arr1.length;
	task.closestZeroSum(arr1, n);
	// Test B
	n = arr2.length;
	task.closestZeroSum(arr2, n);
}
main();

Output

 (7) + (-6) = 1
 (2) + (-2) = 0
#    Python 3 program for
#    Two elements whose sum is closest to zero
class Test :
	def absValue(self, num) :
		if (num < 0) :
			return -num
		
		return num
	
	def closestZeroSum(self, arr, n) :
		if (n <= 1) :
			return
		
		#  Get value of first element
		x = arr[0]
		#  Get value of second element
		y = arr[1]
		sum = x + y
		i = 0
		#  Execute the loop through by size of list
		while (i < n) :
			j = i + 1
			while (j < n) :
				if (self.absValue(sum) > self.absValue(arr[i] + arr[j])) :
					#  Get pair of closest to zero
					x = arr[i]
					y = arr[j]
					#  Change sum
					sum = x + y
				
				j += 1
			
			i += 1
		
		#  Display resultant pair
		print(" (", x ,") + (", y ,") = ", sum)
	

def main() :
	task = Test()
	arr1 = [4, 7, 8, 2, -6, 12]
	arr2 = [2, 6, 8, -2, 5]
	#  Test A
	n = len(arr1)
	task.closestZeroSum(arr1, n)
	#  Test B
	n = len(arr2)
	task.closestZeroSum(arr2, n)

if __name__ == "__main__": main()

Output

 ( 7 ) + ( -6 ) =  1
 ( 2 ) + ( -2 ) =  0
#    Ruby program for
#    Two elements whose sum is closest to zero
class Test 
	def absValue(num) 
		if (num < 0) 
			return -num
		end

		return num
	end

	def closestZeroSum(arr, n) 
		if (n <= 1) 
			return
		end

		#  Get value of first element
		x = arr[0]
		#  Get value of second element
		y = arr[1]
		sum = x + y
		i = 0
		#  Execute the loop through by size of array
		while (i < n) 
			j = i + 1
			while (j < n) 
				if (self.absValue(sum) > self.absValue(arr[i] + arr[j])) 
					#  Get pair of closest to zero
					x = arr[i]
					y = arr[j]
					#  Change sum
					sum = x + y
				end

				j += 1
			end

			i += 1
		end

		#  Display resultant pair
		print(" (", x ,") + (", y ,") = ", sum, "\n")
	end

end

def main() 
	task = Test.new()
	arr1 = [4, 7, 8, 2, -6, 12]
	arr2 = [2, 6, 8, -2, 5]
	#  Test A
	n = arr1.length
	task.closestZeroSum(arr1, n)
	#  Test B
	n = arr2.length
	task.closestZeroSum(arr2, n)
end

main()

Output

 (7) + (-6) = 1
 (2) + (-2) = 0

/*
    Scala program for
    Two elements whose sum is closest to zero
*/
class Test()
{
	def absValue(num: Int): Int = {
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	def closestZeroSum(arr: Array[Int], n: Int): Unit = {
		if (n <= 1)
		{
			return;
		}
		// Get value of first element
		var x: Int = arr(0);
		// Get value of second element
		var y: Int = arr(1);
		var sum: Int = x + y;
		var i: Int = 0;
		// Execute the loop through by size of array
		while (i < n)
		{
			var j: Int = i + 1;
			while (j < n)
			{
				if (absValue(sum) > absValue(arr(i) + arr(j)))
				{
					// Get pair of closest to zero
					x = arr(i);
					y = arr(j);
					// Change sum
					sum = x + y;
				}
				j += 1;
			}
			i += 1;
		}
		// Display resultant pair
		println(" (" + x + ") + (" + y + ") = " + sum);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Test = new Test();
		var arr1: Array[Int] = Array(4, 7, 8, 2, -6, 12);
		var arr2: Array[Int] = Array(2, 6, 8, -2, 5);
		// Test A
		var n: Int = arr1.length;
		task.closestZeroSum(arr1, n);
		// Test B
		n = arr2.length;
		task.closestZeroSum(arr2, n);
	}
}

Output

 (7) + (-6) = 1
 (2) + (-2) = 0
import Foundation;
/*
    Swift 4 program for
    Two elements whose sum is closest to zero
*/
class Test
{
	func absValue(_ num: Int) -> Int
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	func closestZeroSum(_ arr: [Int], _ n: Int)
	{
		if (n <= 1)
		{
			return;
		}
		// Get value of first element
		var x: Int = arr[0];
		// Get value of second element
		var y: Int = arr[1];
		var sum: Int = x + y;
		var i: Int = 0;
		// Execute the loop through by size of array
		while (i < n)
		{
			var j: Int = i + 1;
			while (j < n)
			{
				if (self.absValue(sum) > self.absValue(arr[i] + arr[j]))
				{
					// Get pair of closest to zero
					x = arr[i];
					y = arr[j];
					// Change sum
					sum = x + y;
				}
				j += 1;
			}
			i += 1;
		}
		// Display resultant pair
		print(" (", x ,") + (", y ,") = ", sum);
	}
}
func main()
{
	let task: Test = Test();
	let arr1: [Int] = [4, 7, 8, 2, -6, 12];
	let arr2: [Int] = [2, 6, 8, -2, 5];
	// Test A
	var n: Int = arr1.count;
	task.closestZeroSum(arr1, n);
	// Test B
	n = arr2.count;
	task.closestZeroSum(arr2, n);
}
main();

Output

 ( 7 ) + ( -6 ) =  1
 ( 2 ) + ( -2 ) =  0
/*
    Kotlin program for
    Two elements whose sum is closest to zero
*/
class Test
{
	fun absValue(num: Int): Int
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	fun closestZeroSum(arr: Array < Int > , n: Int): Unit
	{
		if (n <= 1)
		{
			return;
		}
		// Get value of first element
		var x: Int = arr[0];
		// Get value of second element
		var y: Int = arr[1];
		var sum: Int = x + y;
		var i: Int = 0;
		// Execute the loop through by size of array
		while (i < n)
		{
			var j: Int = i + 1;
			while (j < n)
			{
				if (this.absValue(sum) > this.absValue(arr[i] + arr[j]))
				{
					// Get pair of closest to zero
					x = arr[i];
					y = arr[j];
					// Change sum
					sum = x + y;
				}
				j += 1;
			}
			i += 1;
		}
		// Display resultant pair
		println(" (" + x + ") + (" + y + ") = " + sum);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Test = Test();
	val arr1: Array < Int > = arrayOf(4, 7, 8, 2, -6, 12);
	val arr2: Array < Int > = arrayOf(2, 6, 8, -2, 5);
	// Test A
	var n: Int = arr1.count();
	task.closestZeroSum(arr1, n);
	// Test B
	n = arr2.count();
	task.closestZeroSum(arr2, n);
}

Output

 (7) + (-6) = 1
 (2) + (-2) = 0

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