Sum of bitwise AND of all possible subsets of given set
Given an array, we need to calculate the Sum of Bit-wise AND of all possible subsets of given array.
Examples:
Input : 1 2 3
Output : 9
For [1, 2, 3], all possible subsets are {1},
{2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
Bitwise AND of these subsets are, 1 + 2 +
3 + 0 + 1 + 2 + 0 = 9.
So, the answer would be 9.
Input : 1 2 3 4
Output : 13
Refer this Post for Count Set Bit
Naive Approach, we can produce all subset using Power Set then calculate Bit-wise AND sum of all subset.
A Better approach, we are trying to calculate which array element is responsible in producing the sum into subset.
Let’s start with the least significant bit. To remove the contribution from other bits, we calculate number AND bit for all numbers in the set. Any subset of this that contain a 0 will not give any contribution. All nonempty subset that only consist of 1’s will give 1 in contribution. In total there will be 2^n – 1 such subset each giving 1 in contribution. Same goes for the other bit. We get [0, 2, 2], 3 subset each giving 2. Total 3*1 + 3*2 = 9
Array = {1, 2, 3}
Binary representation
positions 2 1 0
1 0 0 1
2 0 1 0
3 0 1 1
[ 0 2 2 ]
Count set bit for each position
[ 0 3 3 ] subset produced by each
position 2^n -1 i.e. n is total sum
for each position [ 0, 3*2^1, 3*2^0 ]
Now calculate the sum by multiplying
the position value i.e 2^0, 2^1 ... .
0 + 6 + 3 = 9
CPP
// C++ program to calculate sum of Bit-wise // and sum of all subsets of an array #include <bits/stdc++.h> using namespace std; #define BITS 32 int andSum(int arr[], int n) { int ans = 0; // assuming representation of each element is // in 32 bit for (int i = 0; i < BITS; i++) { int countSetBits = 0; // iterating array element for (int j = 0; j < n; j++) { // Counting the set bit of array in // ith position if (arr[j] & (1 << i)) countSetBits++; } // counting subset which produce sum when // particular bit position is set. int subset = (1 << countSetBits) - 1; // multiplying every position subset with 2^i // to count the sum. subset = (subset * (1 << i)); ans += subset; } return ans; } // Drivers code int main() { int arr[] = { 1, 2, 3}; int size = sizeof(arr) / sizeof(arr[0]); cout << andSum(arr, size); return 0; } |
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Java
// Java program to calculate sum of Bit-wise // and sum of all subsets of an array class GFG { static final int BITS = 32; static int andSum(int arr[], int n) { int ans = 0; // assuming representation of each // element is in 32 bit for (int i = 0; i < BITS; i++) { int countSetBits = 0; // iterating array element for (int j = 0; j < n; j++) { // Counting the set bit of // array in ith position if ((arr[j] & (1 << i)) != 0) countSetBits++; } // counting subset which produce // sum when particular bit // position is set. int subset = (1 << countSetBits) - 1; // multiplying every position // subset with 2^i to count the // sum. subset = (subset * (1 << i)); ans += subset; } return ans; } // Drivers code public static void main(String args[]) { int arr[] = { 1, 2, 3}; int size = 3; System.out.println (andSum(arr, size)); } } // This code is contributed by Arnab Kundu. |
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Python3
# Python3 program to calculate sum of # Bit-wise and sum of all subsets of # an array BITS = 32; def andSum(arr, n): ans = 0 # assuming representation # of each element is # in 32 bit for i in range(0, BITS): countSetBits = 0 # iterating array element for j in range(0, n) : # Counting the set bit # of array in ith # position if (arr[j] & (1 << i)) : countSetBits = (countSetBits + 1) # counting subset which # produce sum when # particular bit position # is set. subset = ((1 << countSetBits) - 1) # multiplying every position # subset with 2^i to count # the sum. subset = (subset * (1 << i)) ans = ans + subset return ans # Driver code arr = [1, 2, 3] size = len(arr) print (andSum(arr, size)) # This code is contributed by # Manish Shaw (manishshaw1) |
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C#
// C# program to calculate sum of Bit-wise // and sum of all subsets of an array using System; class GFG { static int BITS = 32; static int andSum(int[] arr, int n) { int ans = 0; // assuming representation of each // element is in 32 bit for (int i = 0; i < BITS; i++) { int countSetBits = 0; // iterating array element for (int j = 0; j < n; j++) { // Counting the set bit of // array in ith position if ((arr[j] & (1 << i)) != 0) countSetBits++; } // counting subset which produce // sum when particular bit position // is set. int subset = (1 << countSetBits) - 1; // multiplying every position subset // with 2^i to count the sum. subset = (subset * (1 << i)); ans += subset; } return ans; } // Drivers code static public void Main() { int []arr = { 1, 2, 3}; int size = 3; Console.WriteLine (andSum(arr, size)); } } // This code is contributed by Arnab Kundu. |
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PHP
<?php // PHP program to calculate sum of Bit-wise // and sum of all subsets of an array $BITS = 32; function andSum( $arr, $n) { global $BITS; $ans = 0; // assuming representation // of each element is // in 32 bit for($i = 0; $i < $BITS; $i++) { $countSetBits = 0; // iterating array element for ( $j = 0; $j < $n; $j++) { // Counting the set bit // of array in ith position if ($arr[$j] & (1 << $i)) $countSetBits++; } // counting subset which // produce sum when // particular bit position // is set. $subset = (1 << $countSetBits) - 1; // multiplying every position // subset with 2^i to count // the sum. $subset = ($subset * (1 << $i)); $ans += $subset; } return $ans; } // Driver code $arr = array(1, 2, 3); $size = count($arr); echo andSum($arr, $size); // This code is contributed by anuj_67. ?> |
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Output:
9
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