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What is the time complexity of the following code snippet?
int func(int n) {
if (n <= 0) {
return 0;
}
return 1 + func(n / 2);
}
int result = func(16);
int func(int n) {
if (n <= 0) {
return 0;
}
return 1 + func(n / 2);
}
int result = func(16);
- a)O(n)
- b)O(log n)
- c)O(sqrt(n))
- d)O(n log n)
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
What is the time complexity of the following code snippet?int func(int...
The function is recursively called with n/2 until n <= 0. The number of recursive calls is proportional to the logarithm of n. Hence, the time complexity is O(log n).
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What is the time complexity of the following code snippet?int func(int...
Understanding the Code Snippet
The function `func(int n)` is a recursive function that performs the following:
- **Base Case**: If `n` is less than or equal to 0, it returns 0.
- **Recursive Case**: It returns `1 + func(n / 2)`, which means it calls itself with `n` divided by 2.
Analyzing the Recursion
To determine the time complexity, let's analyze how many times the function will call itself before reaching the base case:
- **Initial Call**: `func(16)`
- **Subsequent Calls**: `func(8)`, `func(4)`, `func(2)`, `func(1)`, and finally `func(0)`, which terminates the recursion.
Number of Recursive Calls
At each step, `n` is halved. Therefore, the number of times we can halve `n` until it reaches 0 is related to the logarithm of `n`:
- The sequence of calls can be expressed as: `n, n/2, n/4, ..., 1`
- The number of calls made is proportional to how many times you can divide `n` by 2, which is `log₂(n)`.
Conclusion
The time complexity of this function is:
- **O(log n)**
Thus, the correct answer is option **B**. This means that as `n` increases, the number of recursive calls increases logarithmically, making this function efficient for large values of `n`.
The function `func(int n)` is a recursive function that performs the following:
- **Base Case**: If `n` is less than or equal to 0, it returns 0.
- **Recursive Case**: It returns `1 + func(n / 2)`, which means it calls itself with `n` divided by 2.
Analyzing the Recursion
To determine the time complexity, let's analyze how many times the function will call itself before reaching the base case:
- **Initial Call**: `func(16)`
- **Subsequent Calls**: `func(8)`, `func(4)`, `func(2)`, `func(1)`, and finally `func(0)`, which terminates the recursion.
Number of Recursive Calls
At each step, `n` is halved. Therefore, the number of times we can halve `n` until it reaches 0 is related to the logarithm of `n`:
- The sequence of calls can be expressed as: `n, n/2, n/4, ..., 1`
- The number of calls made is proportional to how many times you can divide `n` by 2, which is `log₂(n)`.
Conclusion
The time complexity of this function is:
- **O(log n)**
Thus, the correct answer is option **B**. This means that as `n` increases, the number of recursive calls increases logarithmically, making this function efficient for large values of `n`.
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