Check if all the set bits of the binary representation of N are at least K places away


Given numbers N and K, The task to check if all the set bits of the binary representation of N are at least K places away.

Examples:

Input: N = 5, K = 1 
Output: YES
Explanation: 
Binary of 5 is 101.
The 1's are 1 place far from each other.

Input: N = 10, K = 2
Output: NO
Explanation:
Binary of 10 is 1010.
The 1's are not at least 2 places far from each other.

Approach:

  1. Iterate Over all the bits in the binary representation of N and maintain a variable ‘count’ initialize to 0.
  2. Whenever you find a set bit(1), check if count <= K. If not return false.
  3. If you find a unset bit(0), increases the value of count by 1.

Below is the implementation of the above approach.

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// C++ program to check if all the set
// bits of the binary representation
// of N are at least K places away.
#include <bits/stdc++.h>
using namespace std;
  
bool CheckBits(int N, int K)
{
    // Initialize check and count
    // with 0
    int check = 0;
    int count = 0;
  
    for (int i = 31; i >= 0; i--)
    {
  
        // The i-th bit is a set bit
        if ((1 << i) & N)
        {
  
            // This is the first set bit so,
            // start calculating all the
            // distances between consecutive
            // bits from here
            if (check == 0)
            {
                check = 1;
            }
            else
            {
                // If count is less than K 
                // return false
                if (count < K)
                {
                    return false;
                }
            }
            count = 0;
        }
        else
        {
            // Adding the count as the
            // number of zeroes increase
            // between set bits
            count++;
        }
    }
  
    return true;
}
// Driver code
int main()
{
  
    int N = 5;
    int K = 1;
  
    if(CheckBits(N, K))
    {
        cout << "YES";
    }
    else
    {
        cout << "NO";
    }
  
  
  
    return 0;
}

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Output:

YES

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