Given an integer, count set bits in it.

For example,

**Input: **n = -1 (11..1111)

**Output: **The number of set bits in -1 is 32

**Input: **n = 16 (00001000)

**Output: **The number of set bits in 16 is 1

We have discussed a naive solution and Brian Kernighan’s algorithm to count number of set bits in a number in previous post. Both have worst case time complexity of O(log(n)). In this post, a O(1) time solution is discussed.

The idea is to use a **lookup table** to return number of set bits in a number in constant time. An integer in C/C++ usually takes 4 bytes for storage. That means maximum number it can store is 2^{32} – 1. A lookup table for all 2^{32} – 1 integers will be infeasible (Not to forget, we have negative numbers too). The trick is to create an 8-bit (1 byte) version of the table, then iterate over each byte in the integer to be checked and summing the results of the table lookup.

1 byte with all its bits set is 255 in decimal (11111111 in binary) and all bits unset is 0 in decimal (00000000 in binary). So lookup table should be of size 256 (0-255).

In the solution below, we’re using the macros to generate the lookup table. The lookup table will be generated at compile time by preprocessor. The first and last few numbers of the sequence will be

{ 0, 1, 1, 2, 1, . . . , 7, 6, 7, 7, 8 } as

0 has 0 bits

1 has 1 bits

2 has 1 bits

3 has 2 bits

4 has 1 bits

..

..

251 has 7 bits

252 has 6 bits

253 has 7 bits

254 has 7 bits

255 has 8 bits

**C++ implementation –**

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 |
#include <iostream> #include <bitset> using namespace std; // macros to generate the lookup table (at compile-time) #define B2(n) n, n + 1, n + 1, n + 2 #define B4(n) B2(n), B2(n + 1), B2(n + 1), B2(n + 2) #define B6(n) B4(n), B4(n + 1), B4(n + 1), B4(n + 2) #define COUNT_BITS B6(0), B6(1), B6(1), B6(2) // lookup-table to store the number of bits set for each index // in the table. The macro COUNT_BITS generates the table. unsigned int lookup[256] = { COUNT_BITS }; // Function to count number of set bits in n int numOfSetBits(int n) { // print lookup table (number of bits set for integer i) // for (int i = 0; i < 256; i++) // cout << i << " has " << lookup[i] << " bits\n"; // assuming 32-bit(4 byte) integer, break the integer into 8-bit chunks // Note mask used 0xff is 11111111 in binary int count = lookup[n & 0xff] + // consider first 8 bits lookup[(n >> 8) & 0xff] + // consider next 8 bits lookup[(n >> 16) & 0xff] + // consider next 8 bits lookup[(n >> 24) & 0xff]; // consider last 8 bits return count; } // Count set bits using lookup table int main() { int n = -1; cout << n << " in binary is " << bitset<32>(n) << endl; cout << "The number of set bits in " << n << " is " << numOfSetBits(n) << endl; return 0; } |

`Output:`

-1 in binary is 11111111111111111111111111111111

The number of set bits in -1 is 32

**References: **https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetTable

**Thanks for reading.**

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