Construct array with sum of product of same indexed elements in the given array equal to zero

Given an array, arr[] of size N (always even), the task is to construct a new array consisting of N non-zero integers such that the sum of the product of the same indexed elements of arr[] and the new array is equal to 0. If multiple solutions exist, print any one of them.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: -4 -3 2 1 
Explanation:
Sum of product of same indexed array elements of arr[] with the new array = {1 * (-4) + 2 * (-3) + 3 * (2) + 4 * (1)} = 0. 
Therefore, the required output is -4 -3 2 1.

Input: arr[] = {-1, 2, -3, 6, 4}
Output: 1 1 1 1 -1

Approach: The problem can be solved using the Greedy technique. The idea is based on the following observations:

arr[i] * (-arr[i + 1]) + arr[i + 1] * arr[i] = 0  

Follow the steps below to solve the problem:

• Initialize an array, say newArr[] to store the new array elements such that ?(arr[i] * newArr[i]) = 0
• Traverse the given array using variable i and check if i is even or not. If found to be true then newArr[i] = arr[i + 1].
• Otherwise, newArr[i] = -arr[i – 1]
• Finally, print the values of newArr[].

Below is the implementation of the above approach 

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to generate a new array with product
// of same indexed elements with arr[] equal to 0
void constructNewArraySumZero(int arr[], int N)
{
    // Stores sum of same indexed array
    // elements of arr and new array
    int newArr[N];

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // If i is an even number
        if (i % 2 == 0) {

            // Insert arr[i + 1] into
            // the new array newArr[]
            newArr[i] = arr[i + 1];
        }

        else {

            // Insert -arr[i - 1] into
            // the new array newArr[]
            newArr[i] = -arr[i - 1];
        }
    }

    // Print new array elements
    for (int i = 0; i < N; i++) {
        cout << newArr[i] << " ";
    }
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, -5, -6, 8 };
    int N = sizeof(arr) / sizeof(arr[0]);

    constructNewArraySumZero(arr, N);

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Function to generate a new array with
// product of same indexed elements with
// arr[] equal to 0
static void constructNewArraySumZero(int arr[],
                                     int N)
{

    // Stores sum of same indexed array
    // elements of arr and new array
    int newArr[] = new int[N];

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // If i is an even number
        if (i % 2 == 0)
        {

            // Insert arr[i + 1] into
            // the new array newArr[]
            newArr[i] = arr[i + 1];
        }
        else
        {

            // Insert -arr[i - 1] into
            // the new array newArr[]
            newArr[i] = -arr[i - 1];
        }
    }

    // Print new array elements
    for(int i = 0; i < N; i++)
    {
        System.out.print(newArr[i] + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, -5, -6, 8 };
    int N = arr.length;

    constructNewArraySumZero(arr, N);
}
}

// This code is contributed by susmitakundugoaldanga

Python3

# Python3 program to implement
# the above approach

# Function to generate a new array
# with product of same indexed elements
# with arr[] equal to 0
def constructNewArraySumZero(arr, N):

    # Stores sum of same indexed array
    # elements of arr and new array
    newArr = [0] * N

    # Traverse the array
    for i in range(N):

        # If i is an even number
        if (i % 2 == 0):

            # Insert arr[i + 1] into
            # the new array newArr[]
            newArr[i] = arr[i + 1]

        else:

            # Insert -arr[i - 1] into
            # the new array newArr[]
            newArr[i] = -arr[i - 1]

    # Print new array elements
    for i in range(N):
        print(newArr[i] ,
              end = " ")

# Driver code
arr = [1, 2, 3, -5, -6, 8]
N = len(arr)
constructNewArraySumZero(arr, N)

# This code is contributed by divyeshrabadiya07

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to generate a new array with
// product of same indexed elements with
// arr[] equal to 0
static void constructNewArraySumZero(int[] arr,
                                     int N)
{

    // Stores sum of same indexed array
    // elements of arr and new array
    int[] newArr = new int[N];

    // Traverse the array
    for(int i = 0; i < N; i++)
    {

        // If i is an even number
        if (i % 2 == 0)
        {

            // Insert arr[i + 1] into
            // the new array newArr[]
            newArr[i] = arr[i + 1];
        }
        else
        {

            // Insert -arr[i - 1] into
            // the new array newArr[]
            newArr[i] = -arr[i - 1];
        }
    }

    // Print new array elements
    for(int i = 0; i < N; i++)
    {
        Console.Write(newArr[i] + " ");
    }
}

// Driver Code
public static void Main()
{
    int[] arr = { 1, 2, 3, -5, -6, 8 };
    int N = arr.Length;

    constructNewArraySumZero(arr, N);
}
}

// This code is contributed by code_hunt

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to generate a new array with
// product of same indexed elements with
// arr[] equal to 0
function constructNewArraySumZero(arr, N)
{

    // Stores sum of same indexed array
    // elements of arr and new array
    let newArr = [];

    // Traverse the array
    for(let i = 0; i < N; i++)
    {

        // If i is an even number
        if (i % 2 == 0)
        {

            // Insert arr[i + 1] into
            // the new array newArr[]
            newArr[i] = arr[i + 1];
        }
        else
        {

            // Insert -arr[i - 1] into
            // the new array newArr[]
            newArr[i] = -arr[i - 1];
        }
    }

    // Print new array elements
    for(let i = 0; i < N; i++)
    {
        document.write(newArr[i] + " ");
    }
}

    // Driver Code
    let arr = [ 1, 2, 3, -5, -6, 8 ];
    let N = arr.length;

    constructNewArraySumZero(arr, N);

// This code is contributed by souravghosh0416.
</script>

Output

2 -1 -5 -3 8 6 

Time Complexity: O(N)
Auxiliary Space: O(N)