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:
- Initialize an array, brr[], to store the elements satisfying the given conditions.
- Traverse the array arr[]. For every ith 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 conditionsvoid 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 Codeint 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 approachimport java.util.*;class GFG{// Function to construct an array// with given conditionsstatic 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 Codepublic 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 approachfrom math import log# Function to construct an array# with given conditionsdef 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 Codeif __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)
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