Given a number N, the task is to write C program to count the number of 0s and 1s in the binary representation of N.
Examples:
Input: N = 5
Output:
Count of 0s: 1
Count of 1s: 2
Explanation: Binary representation of 5 is “101”.
Input: N = 22
Output:
Count of 0s: 2
Count of 1s: 3
Explanation: Binary representation of 22 is “10110”.
Method 1 – Naive Approach: The idea is to iterate through all bits in the binary representation of N and increment the count of 0s if current bit is ‘0’ else increment the count of 1s.
Below is the implementation of the above approach:
// C program for the above approach #include <stdio.h> // Function to count the number of 0s // and 1s in binary representation of N void count1s0s( int N)
{ // Initialise count variables
int count0 = 0, count1 = 0;
// Iterate through all the bits
while (N > 0) {
// If current bit is 1
if (N & 1) {
count1++;
}
// If current bit is 0
else {
count0++;
}
N = N >> 1;
}
// Print the count
printf ( "Count of 0s in N is %d\n" , count0);
printf ( "Count of 1s in N is %d\n" , count1);
} // Driver Code int main()
{ // Given Number
int N = 9;
// Function Call
count1s0s(N);
return 0;
} |
Count of 0s in N is 2 Count of 1s in N is 2
Time Complexity: O(log N)
Auxiliary Space: O(1)
Method 2 – Recursive Approach: The above approach can also be implemented using Recursion.
Below is the implementation of the above approach:
// C program for the above approach #include <math.h> #include <stdio.h> // Recursive approach to find the // number of set bit in 1 int recursiveCount( int N)
{ // Base Case
if (N == 0) {
return 0;
}
// Return recursively
return (N & 1) + recursiveCount(N >> 1);
} // Function to find 1s complement int onesComplement( int n)
{ // Find number of bits in the
// given integer
int N = floor (log2(n)) + 1;
// XOR the given integer with
// pow(2, N) - 1
return ((1 << N) - 1) ^ n;
} // Function to count the number of 0s // and 1s in binary representation of N void count1s0s( int N)
{ // Initialise the count variables
int count0, count1;
// Function call to find the number
// of set bits in N
count1 = recursiveCount(N);
// Function call to find 1s complement
N = onesComplement(N);
// Function call to find the number
// of set bits in 1s complement of N
count0 = recursiveCount(N);
// Print the count
printf ( "Count of 0s in N is %d\n" , count0);
printf ( "Count of 1s in N is %d\n" , count1);
} // Driver Code int main()
{ // Given Number
int N = 5;
// Function Call
count1s0s(N);
return 0;
} |
Count of 0s in N is 1 Count of 1s in N is 2
Time Complexity: O(log N)
Auxiliary Space: O(1)
Method 3 – Using Brian Kernighan’s Algorithm
We can find the count of set bits using the steps below:
- Initialise count to 0.
- If N > 0, then update N as N & (N – 1) as this will unset the most set bit from the right as shown below:
if N = 10; Binary representation of N = 1010 Binary representation of N - 1 = 1001 ------------------------------------- Logical AND of N and N - 1 = 1000
- Increment the count for the above steps and repeat the above steps until N becomes 0.
To find the count of 0s in the binary representation of N, find the one’s complement of N and find the count of set bits using the approach discussed above.
Below is the implementation of the above approach:
// C program for the above approach #include <math.h> #include <stdio.h> // Function to find 1s complement int onesComplement( int n)
{ // Find number of bits in the
// given integer
int N = floor (log2(n)) + 1;
// XOR the given integer with
// pow(2, N) - 1
return ((1 << N) - 1) ^ n;
} // Function to implement count of // set bits using Brian Kernighan’s // Algorithm int countSetBits( int n)
{ // Initialise count
int count = 0;
// Iterate until n is 0
while (n) {
n &= (n - 1);
count++;
}
// Return the final count
return count;
} // Function to count the number of 0s // and 1s in binary representation of N void count1s0s( int N)
{ // Initialise the count variables
int count0, count1;
// Function call to find the number
// of set bits in N
count1 = countSetBits(N);
// Function call to find 1s complement
N = onesComplement(N);
// Function call to find the number
// of set bits in 1s complement of N
count0 = countSetBits(N);
// Print the count
printf ( "Count of 0s in N is %d\n" , count0);
printf ( "Count of 1s in N is %d\n" , count1);
} // Driver Code int main()
{ // Given Number
int N = 5;
// Function Call
count1s0s(N);
return 0;
} |
Count of 0s in N is 1 Count of 1s in N is 2
Time Complexity: O(log N)
Auxiliary Space: O(1)