Finding the total number of set-bits from 1 to n

Question

Write an algorithm to find F(n) the number of bits set to 1, in all numbers from 1 to n for any given value of n.

Complexity should be O(log n)

Answer 1

this is coded in java...
logic: say number is 34, binary equal-ant is 10010, which can be written as 10000 + 10. 10000 has 4 zeros, so count of all 1's before this number is 2^4(reason below). so count is 2^4 + 2^1 + 1(number it self). so answer is 35.
*for binary number 10000. total combinations of filling 4 places is 2*2*2*2x2(one or zero). So total combinations of ones is 2*2*2*2.

public static int getOnesCount(int number) {
    String binary = Integer.toBinaryString(number);
    return getOnesCount(binary);
}

private static int getOnesCount(String binary) {
    int i = binary.length();

    if (i>0 && binary.charAt(0) == '1') {
        return gePowerof2(i) + getOnesCount(binary.substring(1));
    }else if(i==0)
        return 1;
    else
        return getOnesCount(binary.substring(1));

}
//can be replaced with any default power function
private static int gePowerof2(int i){
    int count = 1;
    while(i>1){
        count*=2;
        i--;
    }
    return count;
}

Answer 2

x = int(input("Any number:\n"))
y = (bin(x))
print(y)
v = len(y) -2
print(v)

Answer 3

Here is the java function

private static int[] findx(int i) {
    //find the biggest power of two number that is smaller than i
    int c = 0;
    int pow2 = 1;
    while((pow2<< 1) <= i) {
        c++;
        pow2 = pow2 << 1;
    }
    return new int[] {c, pow2};
}

public static int TotalBits2(int number) {
    if(number == 0) {
        return 0;
    }
    int[] xAndPow = findx(number);
    int x = xAndPow[0];
    return x*(xAndPow[1] >> 1) + TotalBits2(number - xAndPow[1]) + number - xAndPow[1] + 1;
}

Answer 4

By the way, this question can also be done by the method of lookup table. Precompute the number of set bits from 0-255 and store it. Post that, we can calculate the number of set bits in any number by breaking a given number into two parts of 8 bits each. For each part, we can lookup in the count array formed in the first step. For example, if there is a 16 bit number like,

x = 1100110001011100, here, the number of set bits = number of set bits in the first byte + number of set bits in the second byte. Therefore, for obtaining, first byte,

y = (x & 0xff) z = (x >> 8) & (0xff) total set bits = count[y] + count[z]

This method will run in O(n) as well.

Answer 5

int countSetBits(int n)
{
    n++;
    int powerOf2 = 2;
    int setBitsCount = n/2;
    while (powerOf2 <= n)
    {
        int numbersOfPairsOfZerosAndOnes = n/powerOf2;
        setBitsCount += (numbersOfPairsOfZerosAndOnes/2) * powerOf2;
        setBitsCount += (numbersOfPairsOfZerosAndOnes&1) ? (n%powerOf2) : 0;
        powerOf2 <<= 1;
    }
    return setBitsCount;
}

Please check my article on geeksforgeeks.org for detailed explanation. Below is the link of the article https://www.geeksforgeeks.org/count-total-set-bits-in-all-numbers-from-1-to-n-set-2/

Answer 6

short and sweet!

 public static int countbits(int num){
    int count=0, n;
    while(num > 0){
        n=0;
        while(num >= 1<<(n+1))
            n++;
        num -= 1<<n;
        count += (num + 1 + (1<<(n-1))*n);
    }
    return count;
}//countbis