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Maximum equilibrium sum in an array

Here given code implementation process.

/*
    C program for
    Maximum equilibrium sum in an array
*/
#include <stdio.h>

#include <limits.h>
 // Display given array elements
void printArray(int arr[], int n)
{
	for (int i = 0; i < n; ++i)
	{
		printf(" %d", arr[i]);
	}
}
// Returns the maximum value of given two numbers
int maxValue(int a, int b)
{
	if (a > b)
	{
		return a;
	}
	return b;
}
void findMaximumSum(int arr[], int n)
{
	int sum = 0;
	int prefix = 0;
	int result = INT_MIN;
	// Sum of array elements
	for (int i = 0; i < n; ++i)
	{
		sum += arr[i];
	}
	// Find equilibrium sum
	for (int i = 0; i < n; i++)
	{
		// Prefix sum
		prefix = prefix + arr[i];
		if (prefix == sum)
		{
			result = maxValue(result, prefix);
		}
		// Reduce sum by current array element  
		sum = sum - arr[i];
	}
	printArray(arr, n);
	printf("\n Result : %d \n", result);
}
int main()
{
	// Given array elements
	int arr1[] = {
		6 , 5 , 2 , -5 , 8 , 1 , 4 , -4 , 7
	};
	int arr2[] = {
		3 , -2 , 1 , 4 , 5 , -3
	};
	// Test A
	// Get the number of elements
	int n = sizeof(arr1) / sizeof(arr1[0]);
	//  [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
	// Result 16 [6 + 5 + 2 + -5 + 8]
	findMaximumSum(arr1, n);
	// Test B
	// Get the number of elements
	n = sizeof(arr2) / sizeof(arr2[0]);
	// [ 3, -2, 1, 4] == [4, 5,-3]
	// Result 6 [ 3 + -2 + 1 + 4]
	findMaximumSum(arr2, n);
	return 0;
}

Output

 6 5 2 -5 8 1 4 -4 7
 Result : 16
 3 -2 1 4 5 -3
 Result : 6
/*
    Java Program for
    Maximum equilibrium sum in an array
*/
public class EquilibriumSum
{
	// Display given array elements
	public void printArray(int[] arr, int n)
	{
		for (int i = 0; i < n; ++i)
		{
			System.out.print(" " + arr[i]);
		}
	}
	// Returns the maximum value of given two numbers
	public int maxValue(int a, int b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	public void findMaximumSum(int[] arr, int n)
	{
		int sum = 0;
		int prefix = 0;
		int result = Integer.MIN_VALUE;
		// Sum of array elements
		for (int i = 0; i < n; ++i)
		{
			sum += arr[i];
		}
		// Find equilibrium sum
		for (int i = 0; i < n; i++)
		{
			// Prefix sum
			prefix = prefix + arr[i];
			if (prefix == sum)
			{
				result = maxValue(result, prefix);
			}
			// Reduce sum by current array element  
			sum = sum - arr[i];
		}
		printArray(arr, n);
		System.out.print("\n Result : " + result + " \n");
	}
	public static void main(String[] args)
	{
		EquilibriumSum task = new EquilibriumSum();
		// Given array elements
		int[] arr1 = {
			6 , 5 , 2 , -5 , 8 , 1 , 4 , -4 , 7
		};
		int[] arr2 = {
			3 , -2 , 1 , 4 , 5 , -3
		};
		// Test A
		// Get the number of elements
		int n = arr1.length;
		//  [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
		// Result 16 [6 + 5 + 2 + -5 + 8]
		task.findMaximumSum(arr1, n);
		// Test B
		// Get the number of elements
		n = arr2.length;
		// [ 3, -2, 1, 4] == [4, 5,-3]
		// Result 6 [ 3 + -2 + 1 + 4]
		task.findMaximumSum(arr2, n);
	}
}

Output

 6 5 2 -5 8 1 4 -4 7
 Result : 16
 3 -2 1 4 5 -3
 Result : 6
// Include header file
#include <iostream>
#include <limits.h>

using namespace std;
/*
    C++ Program for
    Maximum equilibrium sum in an array
*/
class EquilibriumSum
{
	public:
		// Display given array elements
		void printArray(int arr[], int n)
		{
			for (int i = 0; i < n; ++i)
			{
				cout << " " << arr[i];
			}
		}
	// Returns the maximum value of given two numbers
	int maxValue(int a, int b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	void findMaximumSum(int arr[], int n)
	{
		int sum = 0;
		int prefix = 0;
		int result = INT_MIN;
		// Sum of array elements
		for (int i = 0; i < n; ++i)
		{
			sum += arr[i];
		}
		// Find equilibrium sum
		for (int i = 0; i < n; i++)
		{
			// Prefix sum
			prefix = prefix + arr[i];
			if (prefix == sum)
			{
				result = this->maxValue(result, prefix);
			}
			// Reduce sum by current array element  
			sum = sum - arr[i];
		}
		this->printArray(arr, n);
		cout << "\n Result : " << result << " \n";
	}
};
int main()
{
	EquilibriumSum *task = new EquilibriumSum();
	// Given array elements
	int arr1[] = {
		6 , 5 , 2 , -5 , 8 , 1 , 4 , -4 , 7
	};
	int arr2[] = {
		3 , -2 , 1 , 4 , 5 , -3
	};
	// Test A
	// Get the number of elements
	int n = sizeof(arr1) / sizeof(arr1[0]);
	//  [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
	// Result 16 [6 + 5 + 2 + -5 + 8]
	task->findMaximumSum(arr1, n);
	// Test B
	// Get the number of elements
	n = sizeof(arr2) / sizeof(arr2[0]);
	// [ 3, -2, 1, 4] == [4, 5,-3]
	// Result 6 [ 3 + -2 + 1 + 4]
	task->findMaximumSum(arr2, n);
	return 0;
}

Output

 6 5 2 -5 8 1 4 -4 7
 Result : 16
 3 -2 1 4 5 -3
 Result : 6
// Include namespace system
using System;
/*
    Csharp Program for
    Maximum equilibrium sum in an array
*/
public class EquilibriumSum
{
	// Display given array elements
	public void printArray(int[] arr, int n)
	{
		for (int i = 0; i < n; ++i)
		{
			Console.Write(" " + arr[i]);
		}
	}
	// Returns the maximum value of given two numbers
	public int maxValue(int a, int b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	public void findMaximumSum(int[] arr, int n)
	{
		int sum = 0;
		int prefix = 0;
		int result = int.MinValue;
		// Sum of array elements
		for (int i = 0; i < n; ++i)
		{
			sum += arr[i];
		}
		// Find equilibrium sum
		for (int i = 0; i < n; i++)
		{
			// Prefix sum
			prefix = prefix + arr[i];
			if (prefix == sum)
			{
				result = this.maxValue(result, prefix);
			}
			// Reduce sum by current array element  
			sum = sum - arr[i];
		}
		this.printArray(arr, n);
		Console.Write("\n Result : " + result + " \n");
	}
	public static void Main(String[] args)
	{
		EquilibriumSum task = new EquilibriumSum();
		// Given array elements
		int[] arr1 = {
			6 , 5 , 2 , -5 , 8 , 1 , 4 , -4 , 7
		};
		int[] arr2 = {
			3 , -2 , 1 , 4 , 5 , -3
		};
		// Test A
		// Get the number of elements
		int n = arr1.Length;
		//  [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
		// Result 16 [6 + 5 + 2 + -5 + 8]
		task.findMaximumSum(arr1, n);
		// Test B
		// Get the number of elements
		n = arr2.Length;
		// [ 3, -2, 1, 4] == [4, 5,-3]
		// Result 6 [ 3 + -2 + 1 + 4]
		task.findMaximumSum(arr2, n);
	}
}

Output

 6 5 2 -5 8 1 4 -4 7
 Result : 16
 3 -2 1 4 5 -3
 Result : 6
package main
import "math"
import "fmt"
/*
    Go Program for
    Maximum equilibrium sum in an array
*/

// Display given array elements
func printArray(arr[] int, n int) {
	for i := 0 ; i < n ; i++ {
		fmt.Print(" ", arr[i])
	}
}
// Returns the maximum value of given two numbers
func maxValue(a, b int) int {
	if a > b {
		return a
	}
	return b
}
func findMaximumSum(arr[] int, n int) {
	var sum int = 0
	var prefix int = 0
	var result int = math.MaxInt64
	// Sum of array elements
	for i := 0 ; i < n ; i++ {
		sum += arr[i]
	}
	// Find equilibrium sum
	for i := 0 ; i < n ; i++ {
		// Prefix sum
		prefix = prefix + arr[i]
		if prefix == sum {
			result = maxValue(result, prefix)
		}
		// Reduce sum by current array element  
		sum = sum - arr[i]
	}
	printArray(arr, n)
	fmt.Print("\n Result : ", result, " \n")
}
func main() {
	
	// Given array elements
	var arr1 = [] int { 6 , 5 , 2 , -5 , 8 , 1 , 4 , -4 , 7 }
	var arr2 = [] int { 3 , -2 , 1 , 4 , 5 , -3 }
	// Test A
	// Get the number of elements
	var n int = len(arr1)
	//  [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
	// Result 16 [6 + 5 + 2 + -5 + 8]
	findMaximumSum(arr1, n)
	// Test B
	// Get the number of elements
	n = len(arr2)
	// [ 3, -2, 1, 4] == [4, 5,-3]
	// Result 6 [ 3 + -2 + 1 + 4]
	findMaximumSum(arr2, n)
}

Output

 6 5 2 -5 8 1 4 -4 7
 Result : 16
 3 -2 1 4 5 -3
 Result : 6
<?php
/*
    Php Program for
    Maximum equilibrium sum in an array
*/
class EquilibriumSum
{
	// Display given array elements
	public	function printArray($arr, $n)
	{
		for ($i = 0; $i < $n; ++$i)
		{
			echo(" ".$arr[$i]);
		}
	}
	// Returns the maximum value of given two numbers
	public	function maxValue($a, $b)
	{
		if ($a > $b)
		{
			return $a;
		}
		return $b;
	}
	public	function findMaximumSum($arr, $n)
	{
		$sum = 0;
		$prefix = 0;
		$result = -PHP_INT_MAX;
		// Sum of array elements
		for ($i = 0; $i < $n; ++$i)
		{
			$sum += $arr[$i];
		}
		// Find equilibrium sum
		for ($i = 0; $i < $n; $i++)
		{
			// Prefix sum
			$prefix = $prefix + $arr[$i];
			if ($prefix == $sum)
			{
				$result = $this->maxValue($result, $prefix);
			}
			// Reduce sum by current array element  
			$sum = $sum - $arr[$i];
		}
		$this->printArray($arr, $n);
		echo("\n Result : ".$result.
			" \n");
	}
}

function main()
{
	$task = new EquilibriumSum();
	// Given array elements
	$arr1 = array(6, 5, 2, -5, 8, 1, 4, -4, 7);
	$arr2 = array(3, -2, 1, 4, 5, -3);
	// Test A
	// Get the number of elements
	$n = count($arr1);
	//  [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
	// Result 16 [6 + 5 + 2 + -5 + 8]
	$task->findMaximumSum($arr1, $n);
	// Test B
	// Get the number of elements
	$n = count($arr2);
	// [ 3, -2, 1, 4] == [4, 5,-3]
	// Result 6 [ 3 + -2 + 1 + 4]
	$task->findMaximumSum($arr2, $n);
}
main();

Output

 6 5 2 -5 8 1 4 -4 7
 Result : 16
 3 -2 1 4 5 -3
 Result : 6
/*
    Node JS Program for
    Maximum equilibrium sum in an array
*/
class EquilibriumSum
{
	// Display given array elements
	printArray(arr, n)
	{
		for (var i = 0; i < n; ++i)
		{
			process.stdout.write(" " + arr[i]);
		}
	}
	// Returns the maximum value of given two numbers
	maxValue(a, b)
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	findMaximumSum(arr, n)
	{
		var sum = 0;
		var prefix = 0;
		var result = -Number.MAX_VALUE;
		// Sum of array elements
		for (var i = 0; i < n; ++i)
		{
			sum += arr[i];
		}
		// Find equilibrium sum
		for (var i = 0; i < n; i++)
		{
			// Prefix sum
			prefix = prefix + arr[i];
			if (prefix == sum)
			{
				result = this.maxValue(result, prefix);
			}
			// Reduce sum by current array element  
			sum = sum - arr[i];
		}
		this.printArray(arr, n);
		process.stdout.write("\n Result : " + result + " \n");
	}
}

function main()
{
	var task = new EquilibriumSum();
	// Given array elements
	var arr1 = [6, 5, 2, -5, 8, 1, 4, -4, 7];
	var arr2 = [3, -2, 1, 4, 5, -3];
	// Test A
	// Get the number of elements
	var n = arr1.length;
	//  [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
	// Result 16 [6 + 5 + 2 + -5 + 8]
	task.findMaximumSum(arr1, n);
	// Test B
	// Get the number of elements
	n = arr2.length;
	// [ 3, -2, 1, 4] == [4, 5,-3]
	// Result 6 [ 3 + -2 + 1 + 4]
	task.findMaximumSum(arr2, n);
}
main();

Output

 6 5 2 -5 8 1 4 -4 7
 Result : 16
 3 -2 1 4 5 -3
 Result : 6
import sys
#    Python 3 Program for
#    Maximum equilibrium sum in an array
class EquilibriumSum :
	#  Display given list elements
	def printArray(self, arr, n) :
		i = 0
		while (i < n) :
			print(" ", arr[i], end = "")
			i += 1
		
	
	#  Returns the maximum value of given two numbers
	def maxValue(self, a, b) :
		if (a > b) :
			return a
		
		return b
	
	def findMaximumSum(self, arr, n) :
		sum = 0
		prefix = 0
		result = -sys.maxsize
		i = 0
		#  Sum of list elements
		while (i < n) :
			sum += arr[i]
			i += 1
		
		i = 0
		#  Find equilibrium sum
		while (i < n) :
			#  Prefix sum
			prefix = prefix + arr[i]
			if (prefix == sum) :
				result = self.maxValue(result, prefix)
			
			#  Reduce sum by current list element  
			sum = sum - arr[i]
			i += 1
		
		self.printArray(arr, n)
		print("\n Result : ", result ," ")
	

def main() :
	task = EquilibriumSum()
	#  Given list elements
	arr1 = [6, 5, 2, -5, 8, 1, 4, -4, 7]
	arr2 = [3, -2, 1, 4, 5, -3]
	#  Test A
	#  Get the number of elements
	n = len(arr1)
	#   [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
	#  Result 16 [6 + 5 + 2 + -5 + 8]
	task.findMaximumSum(arr1, n)
	#  Test B
	#  Get the number of elements
	n = len(arr2)
	#  [ 3, -2, 1, 4] == [4, 5,-3]
	#  Result 6 [ 3 + -2 + 1 + 4]
	task.findMaximumSum(arr2, n)

if __name__ == "__main__": main()

Output

  6  5  2  -5  8  1  4  -4  7
 Result :  16
  3  -2  1  4  5  -3
 Result :  6
#    Ruby Program for
#    Maximum equilibrium sum in an array
class EquilibriumSum 
	#  Display given array elements
	def printArray(arr, n) 
		i = 0
		while (i < n) 
			print(" ", arr[i])
			i += 1
		end

	end

	#  Returns the maximum value of given two numbers
	def maxValue(a, b) 
		if (a > b) 
			return a
		end

		return b
	end

	def findMaximumSum(arr, n) 
		sum = 0
		prefix = 0
		result = -(2 ** (0. size * 8 - 2))
		i = 0
		#  Sum of array elements
		while (i < n) 
			sum += arr[i]
			i += 1
		end

		i = 0
		#  Find equilibrium sum
		while (i < n) 
			#  Prefix sum
			prefix = prefix + arr[i]
			if (prefix == sum) 
				result = self.maxValue(result, prefix)
			end

			#  Reduce sum by current array element  
			sum = sum - arr[i]
			i += 1
		end

		self.printArray(arr, n)
		print("\n Result : ", result ," \n")
	end

end

def main() 
	task = EquilibriumSum.new()
	#  Given array elements
	arr1 = [6, 5, 2, -5, 8, 1, 4, -4, 7]
	arr2 = [3, -2, 1, 4, 5, -3]
	#  Test A
	#  Get the number of elements
	n = arr1.length
	#   [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
	#  Result 16 [6 + 5 + 2 + -5 + 8]
	task.findMaximumSum(arr1, n)
	#  Test B
	#  Get the number of elements
	n = arr2.length
	#  [ 3, -2, 1, 4] == [4, 5,-3]
	#  Result 6 [ 3 + -2 + 1 + 4]
	task.findMaximumSum(arr2, n)
end

main()

Output

 6 5 2 -5 8 1 4 -4 7
 Result : 16 
 3 -2 1 4 5 -3
 Result : 6 

/*
    Scala Program for
    Maximum equilibrium sum in an array
*/
class EquilibriumSum()
{
	// Display given array elements
	def printArray(arr: Array[Int], n: Int): Unit = {
		var i: Int = 0;
		while (i < n)
		{
			print(" " + arr(i));
			i += 1;
		}
	}
	// Returns the maximum value of given two numbers
	def maxValue(a: Int, b: Int): Int = {
		if (a > b)
		{
			return a;
		}
		return b;
	}
	def findMaximumSum(arr: Array[Int], n: Int): Unit = {
		var sum: Int = 0;
		var prefix: Int = 0;
		var result: Int = Int.MinValue;
		var i: Int = 0;
		// Sum of array elements
		while (i < n)
		{
			sum += arr(i);
			i += 1;
		}
		i = 0;
		// Find equilibrium sum
		while (i < n)
		{
			// Prefix sum
			prefix = prefix + arr(i);
			if (prefix == sum)
			{
				result = maxValue(result, prefix);
			}
			// Reduce sum by current array element  
			sum = sum - arr(i);
			i += 1;
		}
		printArray(arr, n);
		print("\n Result : " + result + " \n");
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: EquilibriumSum = new EquilibriumSum();
		// Given array elements
		var arr1: Array[Int] = Array(6, 5, 2, -5, 8, 1, 4, -4, 7);
		var arr2: Array[Int] = Array(3, -2, 1, 4, 5, -3);
		// Test A
		// Get the number of elements
		var n: Int = arr1.length;
		//  [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
		// Result 16 [6 + 5 + 2 + -5 + 8]
		task.findMaximumSum(arr1, n);
		// Test B
		// Get the number of elements
		n = arr2.length;
		// [ 3, -2, 1, 4] == [4, 5,-3]
		// Result 6 [ 3 + -2 + 1 + 4]
		task.findMaximumSum(arr2, n);
	}
}

Output

 6 5 2 -5 8 1 4 -4 7
 Result : 16
 3 -2 1 4 5 -3
 Result : 6
import Foundation;
/*
    Swift 4 Program for
    Maximum equilibrium sum in an array
*/
class EquilibriumSum
{
	// Display given array elements
	func printArray(_ arr: [Int], _ n: Int)
	{
		var i: Int = 0;
		while (i < n)
		{
			print(" ", arr[i], terminator: "");
			i += 1;
		}
	}
	// Returns the maximum value of given two numbers
	func maxValue(_ a: Int, _ b: Int) -> Int
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	func findMaximumSum(_ arr: [Int], _ n: Int)
	{
		var sum: Int = 0;
		var prefix: Int = 0;
		var result: Int = Int.min;
		var i: Int = 0;
		// Sum of array elements
		while (i < n)
		{
			sum += arr[i];
			i += 1;
		}
		i = 0;
		// Find equilibrium sum
		while (i < n)
		{
			// Prefix sum
			prefix = prefix + arr[i];
			if (prefix == sum)
			{
				result = self.maxValue(result, prefix);
			}
			// Reduce sum by current array element  
			sum = sum - arr[i];
			i += 1;
		}
		self.printArray(arr, n);
		print("\n Result : ", result ," ");
	}
}
func main()
{
	let task: EquilibriumSum = EquilibriumSum();
	// Given array elements
	let arr1: [Int] = [6, 5, 2, -5, 8, 1, 4, -4, 7];
	let arr2: [Int] = [3, -2, 1, 4, 5, -3];
	// Test A
	// Get the number of elements
	var n: Int = arr1.count;
	//  [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
	// Result 16 [6 + 5 + 2 + -5 + 8]
	task.findMaximumSum(arr1, n);
	// Test B
	// Get the number of elements
	n = arr2.count;
	// [ 3, -2, 1, 4] == [4, 5,-3]
	// Result 6 [ 3 + -2 + 1 + 4]
	task.findMaximumSum(arr2, n);
}
main();

Output

  6  5  2  -5  8  1  4  -4  7
 Result :  16
  3  -2  1  4  5  -3
 Result :  6
/*
    Kotlin Program for
    Maximum equilibrium sum in an array
*/
class EquilibriumSum
{
	// Display given array elements
	fun printArray(arr: Array < Int > , n: Int): Unit
	{
		var i: Int = 0;
		while (i < n)
		{
			print(" " + arr[i]);
			i += 1;
		}
	}
	// Returns the maximum value of given two numbers
	fun maxValue(a: Int, b: Int): Int
	{
		if (a > b)
		{
			return a;
		}
		return b;
	}
	fun findMaximumSum(arr: Array < Int > , n: Int): Unit
	{
		var sum: Int = 0;
		var prefix: Int = 0;
		var result: Int = Int.MIN_VALUE;
		var i: Int = 0;
		// Sum of array elements
		while (i < n)
		{
			sum += arr[i];
			i += 1;
		}
		i = 0;
		// Find equilibrium sum
		while (i < n)
		{
			// Prefix sum
			prefix = prefix + arr[i];
			if (prefix == sum)
			{
				result = this.maxValue(result, prefix);
			}
			// Reduce sum by current array element  
			sum = sum - arr[i];
			i += 1;
		}
		this.printArray(arr, n);
		print("\n Result : " + result + " \n");
	}
}
fun main(args: Array < String > ): Unit
{
	val task: EquilibriumSum = EquilibriumSum();
	// Given array elements
	val arr1: Array < Int > = arrayOf(6, 5, 2, -5, 8, 1, 4, -4, 7);
	val arr2: Array < Int > = arrayOf(3, -2, 1, 4, 5, -3);
	// Test A
	// Get the number of elements
	var n: Int = arr1.count();
	//  [6, 5, 2, -5, 8] == [8, 1, 4, -4, 7]
	// Result 16 [6 + 5 + 2 + -5 + 8]
	task.findMaximumSum(arr1, n);
	// Test B
	// Get the number of elements
	n = arr2.count();
	// [ 3, -2, 1, 4] == [4, 5,-3]
	// Result 6 [ 3 + -2 + 1 + 4]
	task.findMaximumSum(arr2, n);
}

Output

 6 5 2 -5 8 1 4 -4 7
 Result : 16
 3 -2 1 4 5 -3
 Result : 6
Last updated on by Kalkicode




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