Which of the following best explains how changing i < n to i

Prerequisite: Analysis of Algorithms
1. What is the time, and space complexity of the following code: 
 

Options: 
 

1. O(N * M) time, O(1) space
2. O(N + M) time, O(N + M) space
3. O(N + M) time, O(1) space
4. O(N * M) time, O(N + M) space

Output: 

Explanation: The first loop is O(N) and the second loop is O(M). Since N and M are independent variables, so we can’t say which one is the leading term. Therefore Time complexity of the given problem will be O(N+M).
Since variables size does not depend on the size of the input, therefore  Space Complexity will be constant or O(1)
2. What is the time complexity of the following code: 
 

Options: 
 

1. O(N)
2. O(N*log(N))
3. O(N * Sqrt(N))
4. O(N*N)

Output: 

Explanation: The above code runs total no of times = N + (N – 1) + (N – 2) + … 1 + 0 = N * (N + 1) / 2 = 1/2 * N^2 + 1/2 * N O(N^2) times.

3. What is the time complexity of the following code: 

Options: 
 

1. O(n)
2. O(N log N)
3. O(n^2)
4. O(n^2Logn)

Output: 

Explanation: If you notice, j keeps doubling till it is less than or equal to n. Several times, we can double a number till it is less than n would be log(n). Let’s take the examples here. for n = 16, j = 2, 4, 8, 16 for n = 32, j = 2, 4, 8, 16, 32 So, j would run for O(log n) steps. i runs for n/2 steps. 

So, total steps = O(n/ 2 * log (n)) = O(n*logn)

1. X will always be a better choice for small inputs
2. X will always be a better choice for large inputs
3. Y will always be a better choice for small inputs
4. X will always be a better choice for all inputs

 Output: 

Explanation: In asymptotic analysis, we consider the growth of the algorithm in terms of input size. An algorithm X is said to be asymptotically better than Y if X takes smaller time than y for all input sizes n larger than a value n0 where n0 > 0.
 

5. What is the time complexity of the following code:

Options: 
 

1. O(N)
2. O(Sqrt(N))
3. O(N / 2)
4. O(log N)

Output: 
 

Explanation: We have to find the smallest x such that ‘(N / 2^x )< 1 OR  2^x > N’ 
x = log(N)

6. Which of the following best describes the useful criterion for comparing the efficiency of algorithms?

1. Time
2. Memory
3. Both of the above
4. None of the above

Explanation: Comparing the efficiency of an algorithm depends on the time and memory taken by  an algorithm. The algorithm which runs in lesser time and takes less memory even for a large input size is considered a more efficient algorithm.

7. How is time complexity measured?

1. By counting the number of algorithms in an algorithm.
2. By counting the number of primitive operations performed by the algorithm on a given input size.
3. By counting the size of data input to the algorithm.
4. None of the above

8. What will be the time complexity of the following code?

Output:

Explanation: Because loops for the kn-1 times, so after taking log it becomes logkn.

9. What will be the time complexity of the following code?

Output:

Explanation: First for loop will run for (n) times and another for loop will be run for (n-1) times so overall time will be n(n-1).

10.  Algorithm A and B have a worst-case running time of O(n) and O(logn), respectively. Therefore, algorithm B always runs faster than algorithm A.

Explanation: The Big-O notation provides an asymptotic comparison in the running time of algorithms. For n < n0​​, algorithm A might run faster than algorithm B, for instance.