Counting set bits means determining how many bits in a binary representation of a number are set to 1.
Example
Input: n = 103 (01100111)
Output: 5
Explanation: The binary representation of 103 is 01100111. Hence, the number of set bits is 5.
Input: n = 15 (1111)
Output: 4
Explanation: The binary representation of 15 is 1111. Thus, the number of set bits is 4.
In C++, we have several different methods to count the number of set bits in a number.
Using Bitwise AND with Shift Operator
In this method, we use the bitwise AND operator (&) to check each bit of the number. We process all bits by repeatedly right-shifting the number once after checking whether each one is set.
Code Implementation
C++
// C++ Program to count the number of set bits in a number
// using AND operation and left shifting
#include <iostream>
using namespace std;
int countSetBits(int n) {
int count = 0;
// Runs till the n is greater than 0
while (n) {
// Check the least significant bit
count += n & 1;
// Right shift the number by 1
n >>= 1;
}
return count;
}
int main() {
int n = 103; // Binary: 01100111
// Counting and printing set bits in n
cout << countSetBits(n) << endl;
return 0;
}
Time Complexity: O(log2n), where n is the given number.
Auxiliary Space: O(1)
Using Brian Kernighan’s Algorithm
Brian Kernighan’s algorithm is an efficient method for counting the number of set bits. It works by resetting the rightmost set bit in each iteration in till the number becomes equal to zero. So the number of times we iterate is the number of set bits in it. To reset the rightmost bit, we use the formula
n &= (n - 1);
where n is the given number.
Code Implementation
C++
// C++ Program to count the number of set bits in a number
// using Brian Kernighan's Algorithm
#include <iostream>
using namespace std;
int countSetBits(int n) {
int count = 0;
// Runs till n becomes 0
while (n) {
// Turn off/reset the rightmost set bit
n &= (n - 1);
// Increment the count of set bits
count++;
}
return count;
}
int main() {
int n = 103; // Binary: 01100111
// Counting and printing set bits in n
cout << countSetBits(n) << endl;
return 0;
}
Time Complexity: O(log2n), where n is the given number.
Auxiliary Space: O(1)
The worst case for Brian Kernighan’s algorithm will be when all the bits are set where it will become equal to the first method. But other than that, it will outperform the first method in every case.
Using Lookup Table
In this method, number of set bits for all possible byte values (0–255) are computed and stored in an table for quick lookups.
Code Implementation
C++
// C++ Program to count the number of set bits in a number
// using a lookup table
#include <iostream>
using namespace std;
// Lookup table to store the number of set bits for bytes
unsigned char bitsSetTable256[256];
// Function to initialize the lookup table
void initialize() {
bitsSetTable256[0] = 0;
for (int i = 1; i < 256; i++) {
bitsSetTable256[i] = (i & 1) + bitsSetTable256[i / 2];
}
}
int countSetBits(int n) {
return bitsSetTable256[n & 0xFF] +
bitsSetTable256[(n >> 8) & 0xFF] +
bitsSetTable256[(n >> 16) & 0xFF] +
bitsSetTable256[(n >> 24) & 0xFF];
}
int main() {
// Initialize the lookup table
initialize();
int n = 103; // Binary: 01100111
// Counting and printing set bits in n
cout << countSetBits(n) << endl;
return 0;
}
Time Complexity: O(1), time for creating lookup table is not considered.
Auxiliary Space: O(1), space for lookup table not considered.
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