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: 2
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
// 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 arrayint 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 Codeint 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 |
Java
// Java implementation to check// if a number divided by power// of two exist in the sorted arrayimport 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)); }} |
Python3
# 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 arraydef 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 Codearr = [ 3, 5, 7, 8, 10 ]k = 4n = 5print(findNumberDivByPowerofTwo(arr, k, n))# This code is contributed by code_hunt |
C#
// C# implementation to check// if a number divided by power// of two exist in the sorted arrayusing System;class GFG{ // Function to find there exist a// number or not in the arraystatic 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 Codepublic 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 |
Javascript
<script>// javascript 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 function findNumberDivByPowerofTwo(ar , k , n) { var found = -1, m = k; // Loop to check if there exist // a number by divided by power of 2 while (m > 0) { var l = 0; var r = n - 1; // Binary Search while (l <= r) { var mid = parseInt((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 = parseInt(m / 2); } return found; } // Driver Code var arr = [ 3, 5, 7, 8, 10 ]; var k = 4, n = 5; document.write(findNumberDivByPowerofTwo(arr, k, n));// This code contributed by gauravrajput1 </script> |
Output:
-1
Time Complexity: O(logn), required to do binary search operations
Auxiliary Space: O(1), as no extra space is used
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