# 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 ith 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 ith 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:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `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

 ``

Output:

`8 4 4 2 2`

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

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