What will be the order of complexity with respect to time of the following function ?

int func(int n)

{int count = 0;

for (int i = n, i > 0; i/= 2)

for (int j = 0; j<i; j++)

count + = 1;

return count;

}

The correct answer is: option 3: O(n)

Concept:

Let’s analyze the time complexity of the given function step-by-step:

int func(int n) {
    int count = 0;
    for (int i = n; i > 0; i /= 2)           / Outer loop: runs log₂n times
        for (int j = 0; j < i; j++)          / Inner loop: runs i times each iteration
            count += 1;
    return count;
}

Outer Loop: i = n; i > 0; i /= 2

• This loop runs ⌊log₂(n)⌋ + 1 times because i is halved in each iteration.

Inner Loop: for (int j = 0; j < i; j++)

• In the first iteration, i = n → inner runs n times
• Next, i = n/2 → inner runs n/2 times
• Then n/4, n/8, ..., up to 1

Total work done =

\( n + n/2 + n/4 + n/8 + \dots \approx 2n \)

Hence, the overall time complexity is O(n).

Explanation of options:

• Option 1 – O(n²): ❌ Too large, not accurate.
• Option 2 – O(nlogn): ❌ Common but incorrect here.
• Option 3 – O(n): ✅ Correct, as explained.
• Option 4 – O(nlogn): ❌ Repetition of option 2, still incorrect.

Hence, the correct answer is: option 3: O(n)