Generate an N-length array with sum equal to twice the sum of its absolute difference with same-indexed elements of given array

Given an array arr[] of size N, the task is to construct an array brr[] of size N that satisfies the following conditions:

• In every pair of consecutive elements in the array brr[], one element must be divisible by the other, i.e. brr[i] must be divisible by brr[i + 1] or vice-versa.
• Every i^th element in the array brr[] must satisfy brr[i] >= arr[i] / 2.
• The sum of elements of the array arr[] must be greater than or equal to 2 * ?abs(arr[i] – brr[i]).

Examples:

Input: arr[] = { 11, 5, 7, 3, 2 } 
Output: 8 4 4 2 2 
Explanation:
abs(11 – 8) + abs(5 – 4) + abs(7 – 4) + abs(3 – 2) + abs(2 – 2) = 8 
arr[0] + arr[1] + … + arr[4] = 28 
2 * 8 <= 28 and for every i^th element brr[i] >= arr[i] / 2. 
Therefore, one of the possible values of brr[] are 8 4 4 2 2.

Input: arr[] = { 11, 7, 5 } 
Output: { 8, 4, 4 }

Approach: The idea is based on the following observation:

If brr[i] is the nearest power of 2 and is smaller than or equal to arr[i], then brr[i] must be greater than or equal to arr[i] / 2 and also the sum of elements of the array, arr[] must be greater than or equal to 2 * ?abs(arr[i] – brr[i]).

Follow the steps below to solve the problem:

• Initialize an array, brr[], to store the elements satisfying the given conditions.
• Traverse the array arr[]. For every i^th element, find the nearest power of 2 which is smaller than or equal to arr[i] and store it in brr[i].
• Finally, print the array brr[].

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to construct an array
// with given conditions
void constructArray(int arr[], int N)
{
    int brr[N] = { 0 };

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

        int K = log(arr[i]) / log(2);

        // Stores closest power of 2
        // less than or equal to arr[i]
        int R = pow(2, K);

        // Stores R into brr[i]
        brr[i] = R;
    }

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

// Driver Code
int main()
{

    // Given array
    int arr[] = { 11, 5, 7, 3, 2 };

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    constructArray(arr, N);

    return 0;
}

Java

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

class GFG{

// Function to construct an array
// with given conditions
static void constructArray(int arr[], int N)
{
    int brr[] = new int[N];

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
        int K = (int)(Math.log(arr[i]) /
                      Math.log(2));

        // Stores closest power of 2
        // less than or equal to arr[i]
        int R = (int)Math.pow(2, K);

        // Stores R into brr[i]
        brr[i] = R;
    }

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

// Driver Code
public static void main(String[] args)
{

    // Given array
    int arr[] = { 11, 5, 7, 3, 2 };

    // Size of the array
    int N = arr.length;

    // Function Call
    constructArray(arr, N);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
from math import log

# Function to construct an array
# with given conditions
def constructArray(arr, N):
    brr = [0]*N

    # Traverse the array arr[]
    for i in range(N):
        K = int(log(arr[i])/log(2))

        # Stores closest power of 2
        # less than or equal to arr[i]
        R = pow(2, K)

        # Stores R into brr[i]
        brr[i] = R

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

# Driver Code
if __name__ == '__main__':

  # Given array
    arr = [11, 5, 7, 3, 2]

    # Size of the array
    N = len(arr)

    # Function Call
    constructArray(arr, N)

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;

class GFG{

// Function to construct an array
// with given conditions
static void constructArray(int []arr, int N)
{
    int[] brr = new int[N];
    Array.Clear(brr, 0, brr.Length);

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {
        int K = (int)(Math.Log(arr[i]) /
                      Math.Log(2));

        // Stores closest power of 2
        // less than or equal to arr[i]
        int R = (int)Math.Pow(2, K);

        // Stores R into brr[i]
        brr[i] = R;
    }

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

// Driver Code
public static void Main()
{

    // Given array
    int []arr = { 11, 5, 7, 3, 2 };

    // Size of the array
    int N = arr.Length;

    // Function Call
    constructArray(arr, N);
}
}

// This code is contributed by SURENDRA_GANGWAR

Javascript

<script>

// JavaScript program for the above approach

    // Function to construct an array
    // with given conditions
    function constructArray(arr , N) {
        var brr = Array(N).fill(0);

        // Traverse the array arr
        for (i = 0; i < N; i++) {
            var K = parseInt( (Math.log(arr[i]) / Math.log(2)));

            // Stores closest power of 2
            // less than or equal to arr[i]
            var R = parseInt( Math.pow(2, K));

            // Stores R into brr[i]
            brr[i] = R;
        }

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

    // Driver Code


        // Given array
        var arr = [ 11, 5, 7, 3, 2 ];

        // Size of the array
        var N = arr.length;

        // Function Call
        constructArray(arr, N);

// This code is contributed by todaysgaurav

</script>

Output:

8 4 4 2 2

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