Check if ceil of number divided by power of two exist in sorted array

Given a sorted array arr[] and an integer K, the task is to check if there exists a ceil of number K divided by some power of 2 in the array.
Note: If there is no such element print -1. 
Examples:

Input: arr[] = {3, 5, 7, 8, 10}, K = 4 
Output: -1 
Explanation: 
There is no such element.

Input: arr[] = {1, 2, 3, 5, 7, 8}, K = 4 
Output:
Explanation: 
When 4 is divided by 2 to the power 1, there exist an element in the array.

Approach: The idea is to try for every power of 2 and check that there exists a ceil of that number starting from power as 0. Checking an element exists in the array or not can be done using Binary Search.

Below is the implementation of the above approach:

C++14

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// C++14 implementation to check
// if a number divided by power
// of two exist in the sorted array
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find there exist a
// number or not in the array
int findNumberDivByPowerofTwo(int ar[], 
                              int k, int n)
{
    int found = -1, m = k;
  
    // Loop to check if there exist
    // a number by divided by power of 2
    while (m > 0)
    {
        int l = 0;
        int r = n - 1;
  
        // Binary Search
        while (l <= r)
        {
            int mid = (l + r) / 2;
  
            if (ar[mid] == m)
            {
                found = m;
                break;
            }
            else if (ar[mid] > m)
            {
                r = mid - 1;
            }
            else if (ar[mid] < m)
            {
                l = mid + 1;
            }
        }
  
        // Condition to check the number
        // is found in the array or not
        if (found != -1)
        {
            break;
        }
  
        // Otherwise divide the number
        // by increasing the one more
        // power of 2
        m = m / 2;
    }
    return found;
}
  
// Driver Code
int main()
{
    int arr[] = { 3, 5, 7, 8, 10 };
    int k = 4, n = 5;
      
    cout << findNumberDivByPowerofTwo(arr, k, n);
}
  
// This code is contributed by code_hunt

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Java

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// Java implementation to check
// if a number divided by power
// of two exist in the sorted array
  
import java.util.Scanner;
  
public class GreeksForGreeksQuestions {
  
    // Function to find there exist a
    // number or not in the array
    static int findNumberDivByPowerofTwo(
        int[] ar, int k, int n)
    {
        int found = -1, m = k;
  
        // Loop to check if there exist
        // a number by divided by power of 2
        while (m > 0) {
            int l = 0;
            int r = n - 1;
  
            // Binary Search
            while (l <= r) {
                int mid = (l + r) / 2;
  
                if (ar[mid] == m) {
                    found = m;
                    break;
                }
                else if (ar[mid] > m) {
                    r = mid - 1;
                }
                else if (ar[mid] < m) {
                    l = mid + 1;
                }
            }
  
            // Condition to check the number
            // is found in the array or not
            if (found != -1) {
                break;
            }
  
            // Otherwise divide the number
            // by increasing the one more
            // power of 2
            m = m / 2;
        }
  
        return found;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 3, 5, 7, 8, 10 };
        int k = 4, n = 5;
  
        System.out.println(
            findNumberDivByPowerofTwo(
                arr, k, n));
    }
}

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Python3

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# Python3 implementation to check
# if a number divided by power
# of two exist in the sorted array
  
# Function to find there exist a
# number or not in the array
def findNumberDivByPowerofTwo(ar, k, n):
      
    found = -1
    m = k
  
    # Loop to check if there exist
    # a number by divided by power of 2
    while (m > 0):
        l = 0
        r = n - 1
  
        # Binary Search
        while (l <= r):
            mid = (l + r) // 2
  
            if (ar[mid] == m):
                found = m
                break
              
            elif (ar[mid] > m):
                r = mid - 1
              
            elif (ar[mid] < m):
                l = mid + 1
              
        # Condition to check the number
        # is found in the array or not
        if (found != -1):
            break
          
        # Otherwise divide the number
        # by increasing the one more
        # power of 2
        m = m // 2
      
    return found
  
# Driver Code
arr = [ 3, 5, 7, 8, 10 ]
k = 4
n = 5
  
print(findNumberDivByPowerofTwo(arr, k, n))
  
# This code is contributed by code_hunt

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C#

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// C# implementation to check
// if a number divided by power
// of two exist in the sorted array
using System;
  
class GFG{
      
// Function to find there exist a
// number or not in the array
static int findNumberDivByPowerofTwo(int[] ar, int k,
                                     int n)
{
    int found = -1, m = k;
  
    // Loop to check if there exist
    // a number by divided by power of 2
    while (m > 0) 
    {
        int l = 0;
        int r = n - 1;
  
        // Binary Search
        while (l <= r)
        {
            int mid = (l + r) / 2;
  
            if (ar[mid] == m)
            {
                found = m;
                break;
            }
            else if (ar[mid] > m)
            {
                r = mid - 1;
            }
            else if (ar[mid] < m)
            {
                l = mid + 1;
            }
        }
  
        // Condition to check the number
        // is found in the array or not
        if (found != -1)
        {
            break;
        }
  
        // Otherwise divide the number
        // by increasing the one more
        // power of 2
        m = m / 2;
    }
    return found;
}
  
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 3, 5, 7, 8, 10 };
    int k = 4, n = 5;
  
    Console.WriteLine(findNumberDivByPowerofTwo(
                      arr, k, n));
}
}
  
// This code is contributed by princi singh

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Output: 

-1

 

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Improved By : princi singh, code_hunt