Number of integers with odd number of set bits in C++

In this tutorial, we are going to write a program that counts the number of integers with odd number of set bits.

Let's see the steps to solve the problem.

Initialise the number N.
Initialise the count to 0.
Write a loop that iterates from 1 to N.
- Count the set bits of each integer.
- Increment the count if the set bits count is odd.
Return the count.

Example

Let's see the code.

#include <bits/stdc++.h>
using namespace std;
int getSetBitsCount(int n) {
   int count = 0;
   while (n) {
      if (n % 2 == 1) {
         count += 1;
      }
      n /= 2;
   }
   return count;
}
int getOddSetBitsIntegerCount(int n) {
   int count = 0;
   for (int i = 1; i <= n; i++) {
      if (getSetBitsCount(i) % 2 == 1) {
         count += 1;
      }
   }
   return count;
}
int main() {
   int n = 10;
   cout << getOddSetBitsIntegerCount(n) << endl;
   return 0;
}

Output

If you run the above code, then you will get the following result.

Conclusion

If you have any queries in the tutorial, mention them in the comment section.

Hafeezul Kareem

Published on 03-Jul-2021 05:25:47

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