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**element in the array^{th}**brr[]**must satisfy**brr[i] >= arr[i] / 2**. - The sum of elements of the array
**arr[]**must be greater than or equal to**2 * Σabs(arr[i] – brr[i])**.

**Examples:**

Input:arr[] = { 11, 5, 7, 3, 2 }Output:8 4 4 2 2Explanation:

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 toarr[i], thenbrr[i]must be greater than or equal toarr[i] / 2and also the sum of elements of the array,arr[]must be greater than or equal to2 * Σabs(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**element, find the nearest power of^{th}**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;` `}` |

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## Java

`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to conan 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` |

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## 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` ` ` `# Prarray 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` |

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## 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` |

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**Output:**

8 4 4 2 2

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

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