Given an array arr[] of n positive integers. The task is to check if there exist any subset of the array whose bitwise AND is a power of two (i.e 1, 2, 4, 8, 16, …).
Examples:
Input : n = 3, arr[] = { 12, 13, 7 }
Output : Yes
Subset { 12, 7 } has Biwise AND value 4, which
is power of 2.
Input : n = 2, arr[] = { 10, 20 }
Output : No
Observe, for a number to be the power of 2, it should have only 1 set bit.
If n is 1, then we simply check if the number has the only single set bit.
For n is greater than one, our task becomes to choose those numbers from the array whose bitwise AND leads to an only single bit set number. To do so, we search a position, at which all elements in the set has a bit set at that position. For example, for set { 4 (100), 6 (110), 7 (111) }, at position 2 (from right to left, 0-based indexing) bit is set for all element. So, doing bitwise AND gives 4, which is a power of 2.
Below is the implementation of this approach:
C++
// CPP Program to check if Bitwise AND of any // subset is power of two #include <bits/stdc++.h> using namespace std; const int NUM_BITS = 32; // Check for power of 2 or not bool isPowerOf2(int num) { return (num && !(num & (num - 1))); } // Check if there exist a subset whose bitwise AND // is power of 2. bool checkSubsequence(int arr[], int n) { // if there is only one element in the set. if (n == 1) return isPowerOf2(arr[0]); // Finding a number with all bit sets. int total = 0; for (int i = 0; i < NUM_BITS; i++) total = total | (1 << i); // check all the positions at which the bit is set. for (int i = 0; i < NUM_BITS; i++) { int ans = total; for (int j = 0; j < n; j++) { // include all those elements whose // i-th bit is set if (arr[j] & (1 << i)) ans = ans & arr[j]; } // check for the set contains elements // make a power of 2 or not if (isPowerOf2(ans)) return true; } return false; } // Driver Program int main() { int arr[] = { 12, 13, 7 }; int n = sizeof(arr) / sizeof(arr[0]); if (checkSubsequence(arr, n)) printf("YES\n"); else printf("NO\n"); return 0; } |
chevron_right
filter_none
Java
// Java Program to check if Bitwise AND of any // subset is power of two import java.io.*; import java.util.*; public class GFG { static int NUM_BITS = 32; // Check for power of 2 or not static boolean isPowerOf2(int num) { if(num != 0 && (num & (num - 1)) == 0) return true; return false; } // Check if there exist a // subset whose bitwise AND // is power of 2. static boolean checkSubsequence(int []arr, int n) { // if there is only one // element in the set. if (n == 1) return isPowerOf2(arr[0]); // Finding a number with // all bit sets. int total = 0; for (int i = 0; i < NUM_BITS; i++) total = total | (1 << i); // check all the positions // at which the bit is set. for (int i = 0; i < NUM_BITS; i++) { int ans = total; for (int j = 0; j < n; j++) { // include all those // elements whose // i-th bit is set int p = arr[j] & (1 << i); if (p == 0) ans = ans & arr[j]; } // check for the set // contains elements // make a power of 2 // or not if (isPowerOf2(ans)) return true; } return false; } // Driver Code public static void main(String args[]) { int []arr = {12, 13, 7}; int n = arr.length; if (checkSubsequence(arr, n)) System.out.println("YES"); else System.out.println("NO"); } } // This code is contributed by // Manish Shaw (manishshaw1) |
chevron_right
filter_none
Python3
# Python3 Program to check if Bitwise AND of any # subset is power of two NUM_BITS = 32 # Check for power of 2 or not def isPowerOf2(num): return (num and (num & (num - 1)) == 0) # Check if there exist a subset whose bitwise AND # is power of 2. def checkSubsequence(arr, n): # if there is only one element in the set. if (n == 1): return isPowerOf2(arr[0]) # Finding a number with all bit sets. total = 0 for i in range(0, NUM_BITS): total = total | (1 << i) # check all the positions at which the bit is set. for i in range(0, NUM_BITS): ans = total for j in range(0, n): # include all those elements whose # i-th bit is set if (arr[j] & (1 << i)): ans = ans & arr[j] # check for the set contains elements # make a power of 2 or not if (isPowerOf2(ans)): return True return False # Driver Program arr = [ 12, 13, 7 ] n = len(arr) if (checkSubsequence(arr, n)): print ("YES\n") else: print ("NO\n") # This code is contributed by Manish Shaw # (manishshaw1) |
chevron_right
filter_none
C#
// C# Program to check if Bitwise AND of any // subset is power of two using System; using System.Collections.Generic; class GFG { static int NUM_BITS = 32; // Check for power of 2 or not static bool isPowerOf2(int num) { if(num != 0 && (num & (num - 1)) == 0) return true; return false; } // Check if there exist a // subset whose bitwise AND // is power of 2. static bool checkSubsequence(int []arr, int n) { // if there is only one // element in the set. if (n == 1) return isPowerOf2(arr[0]); // Finding a number with // all bit sets. int total = 0; for (int i = 0; i < NUM_BITS; i++) total = total | (1 << i); // check all the positions // at which the bit is set. for (int i = 0; i < NUM_BITS; i++) { int ans = total; for (int j = 0; j < n; j++) { // include all those // elements whose // i-th bit is set int p = arr[j] & (1 << i); if (p == 0) ans = ans & arr[j]; } // check for the set // contains elements // make a power of 2 // or not if (isPowerOf2(ans)) return true; } return false; } // Driver Code public static void Main() { int []arr = {12, 13, 7}; int n = arr.Length; if (checkSubsequence(arr, n)) Console.Write("YES\n"); else Console.Write("NO\n"); } } // This code is contributed by // Manish Shaw (manishshaw1) |
chevron_right
filter_none
PHP
<?php // PHP Program to check if // Bitwise AND of any subset // is power of two // Check for power of 2 or not function isPowerOf2($num) { return ($num && !($num & ($num - 1))); } // Check if there exist a // subset whose bitwise AND // is power of 2. function checkSubsequence($arr, $n) { $NUM_BITS = 32; // if there is only one // element in the set. if ($n == 1) return isPowerOf2($arr[0]); // Finding a number with // all bit sets. $total = 0; for($i = 0; $i < $NUM_BITS; $i++) $total = $total | (1 << $i); // check all the positions at // which the bit is set. for($i = 0; $i < $NUM_BITS; $i++) { $ans = $total; for ($j = 0; $j < $n; $j++) { // include all those // elements whose // i-th bit is set if ($arr[$j] & (1 << $i)) $ans = $ans & $arr[$j]; } // check for the set // contains elements // make a power of 2 or not if (isPowerOf2($ans)) return true; } return false; } // Driver Code $arr= array(12, 13, 7); $n = sizeof($arr) / sizeof($arr[0]); if (checkSubsequence($arr, $n)) echo "YES"; else echo "NO"; // This code is contributed by mits ?> |
chevron_right
filter_none
Output:
YES
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Check whether bitwise AND of a number with any subset of an array is zero or not
- Total pairs in an array such that the bitwise AND, bitwise OR and bitwise XOR of LSB is 1
- Minimum possible Bitwise OR of all Bitwise AND of pairs generated from two given arrays
- Sum of maximum and minimum of Kth subset ordered by increasing subset sum
- Count ways to generate pairs having Bitwise XOR and Bitwise AND equal to X and Y respectively
- Check if given number is a power of d where d is a power of 2
- Minimum Bitwise AND operations to make any two array elements equal
- Find subsequences with maximum Bitwise AND and Bitwise OR
- Maximize sum of squares of array elements possible by replacing pairs with their Bitwise AND and Bitwise OR
- Count pairs with equal Bitwise AND and Bitwise OR value
- Maximum subset with bitwise OR equal to k
- Size of the smallest subset with maximum Bitwise OR
- Minimum Bitwise OR operations to make any two array elements equal
- Minimum Bitwise XOR operations to make any two array elements equal
- Leftover element after performing alternate Bitwise OR and Bitwise XOR operations on adjacent pairs
- Count pairs with bitwise XOR exceeding bitwise AND from a given array
- Count of pairs whose bitwise AND is a power of 2
- Find if there is any subset of size K with 0 sum in an array of -1 and +1
- Find the sum of power of bit count raised to the power B
- Find the smallest positive integer value that cannot be represented as sum of any subset of a given array
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
