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# C program to count zeros and ones in binary representation of a number

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

 `// C program for the above approach` `#include `   `// 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;` `}`

Output

```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

 `// C program for the above approach` `#include ` `#include `   `// 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;` `}`

Output

```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

 `// C program for the above approach` `#include ` `#include `   `// 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;` `}`

Output

```Count of 0s in N is 1
Count of 1s in N is 2```

Time Complexity: O(log N)
Auxiliary Space: O(1)

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