Test: Time and Space Complexity - 2 - Software Development MCQ

The O-notation represents the worst-case time complexity of an algorithm. It indicates that the time taken by the algorithm grows linearly with the input size.

O(log n) represents logarithmic time complexity. It indicates that the time taken by the algorithm increases slowly as the input size grows.

O-notation represents the upper bound of the time complexity. It provides an upper limit on the growth rate of an algorithm.

O(log n) grows slower than O(sqrt(n)). Therefore, O(log n) is faster in terms of time complexity.

In the given equation, 5n^2 has the highest power of n. Hence, it is the dominant term.

The outer loop runs n times, and the inner loop runs log(n) times. Therefore, the total number of iterations is n * log(n), resulting in a time complexity of O(log n).

The function recursively calls itself twice for each value. Since the input is halved at each step, the total number of recursive calls is 2^3 = 8. Therefore, the output is 2^8 = 256.

The given code divides n by 2 repeatedly until n becomes 0. The number of divisions required is log2(n) + 1. Therefore, for n = 100, log2(100) + 1 = 6 + 1 = 7.

The outer loop iterates from 1 to n. For each value of i, the inner loop iterates from i to n, doubling the value of j at each iteration. The output is printed after each inner loop.

The outer loop starts with n and is divided by 2 in each iteration until it becomes 1. The number of outer loop iterations is log2(n) + 1. Therefore, for n = 4, log2(4) + 1 = 2 + 1 = 3. The inner loop runs 1, 2, 4 times. Hence, the total count is 3 * 4 = 12.

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).

The function func is recursively called twice for each value of n. The maximum depth of the recursion is n. Therefore, the space complexity is O(n).

The statement is false. The time complexity of an algorithm can vary based on the input size and the algorithm's implementation. It is not always constant.

The outer loop runs n times, and the inner loop doubles the value of j in each iteration until it reaches n. Therefore, the total number of iterations is n * log(n). Hence, the time complexity is O(n^2).

The given code implements the Fibonacci sequence using recursion. It recursively calls the function twice for each value until n <= 1. The number of function calls grows exponentially with n. Hence, the space complexity is O(2^n).