[02 Aug 2020] Find position of the only set bit

 #include <bits/stdc++.h>

using namespace std;

// A utility function to check whether n is a power of 2 or not.
// See http://goo.gl/17Arj
int isPowerOfTwo(unsigned n)
{
    return n && (!(n & (n - 1)));
}
  
// Returns position of the only set bit in 'n'
int findPosition(unsigned n)
{
    if (!isPowerOfTwo(n))
        return -1;
  
    unsigned i = 1, pos = 1;
  
    // Iterate through bits of n till we find a set bit
    // i&n will be non-zero only when 'i' and 'n' have a set bit
    // at same position
    while (!(i & n)) {
        // Unset current bit and set the next bit in 'i'
        i = i << 1;
  
        // increment position
        ++pos;
    }
  
    return pos;
}
  
// Driver program to test above function
int main(void)
{
    int n = 16;
    int pos = findPosition(n);
    (pos == -1) ? cout << "n = " << n << ", Invalid number" << endl : cout << "n = " << n << ", Position " << pos << endl;
  
    n = 12;
    pos = findPosition(n);
    (pos == -1) ? cout << "n = " << n << ", Invalid number" << endl : cout << "n = " << n << ", Position " << pos << endl;
  
    n = 128;
    pos = findPosition(n);
    (pos == -1) ? cout << "n = " << n << ", Invalid number" << endl : cout << "n = " << n << ", Position " << pos << endl;
  
    return 0;
}

[2] Count total set bits in all numbers from 1 to n


Given a positive integer n, count the total number of set bits in binary representation of all numbers from 1 to n.

Examples:

Input: n = 3
Output:  4

Input: n = 6
Output: 9

Input: n = 7
Output: 12

Input: n = 8
Output: 13
// A simple program to count set bits
// in all numbers from 1 to n.
#include <stdio.h>
  
// A utility function to count set bits
// in a number x
unsigned int countSetBitsUtil(unsigned int x);
  
// Returns count of set bits present in all
// numbers from 1 to n
unsigned int countSetBits(unsigned int n)
{
    int bitCount = 0; // initialize the result
  
    for (int i = 1; i <= n; i++)
        bitCount += countSetBitsUtil(i);
  
    return bitCount;
}
  
// A utility function to count set bits 
// in a number x
unsigned int countSetBitsUtil(unsigned int x)
{
    if (x <= 0)
        return 0;
    return (x % 2 == 0 ? 0 : 1) + countSetBitsUtil(x / 2);
}
  
// Driver program to test above functions
int main()
{
    int n = 4;
    printf("Total set bit count is %d", countSetBits(n));
    return 0;
} 

Total set bit count is 5

Time Complexity: O(nLogn)








































































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