Prerequiste: Analysis of Algorithms
1. What is the time, space complexity of following code:
int a = 0, b = 0; for (i = 0; i < N; i++) { a = a + rand(); } for (j = 0; j < M; j++) { b = b + rand(); } |
Options:
- O(N * M) time, O(1) space
- O(N + M) time, O(N + M) space
- O(N + M) time, O(1) space
- O(N * M) time, O(N + M) space
Output:
3. O(N + M) time, O(1) space
Explanation: The first loop is O(N) and the second loop is O(M). Since we don’t know which is bigger, we say this is O(N + M). This can also be written as O(max(N, M)).
Since there is no additional space being utilized, the space complexity is constant / O(1)
2. What is the time complexity of following code:
int a = 0; for (i = 0; i < N; i++) { for (j = N; j > i; j--) { a = a + i + j; } } |
Options:
- O(N)
- O(N*log(N))
- O(N * Sqrt(N))
- O(N*N)
Output:
4. O(N*N)
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 following code:
int i, j, k = 0; for (i = n / 2; i <= n; i++) { for (j = 2; j <= n; j = j * 2) { k = k + n / 2; } } |
Options:
- O(n)
- O(nLogn)
- O(n^2)
- O(n^2Logn)
Output:
2. O(nLogn)
Explanation:If you notice, j keeps doubling till it is less than or equal to n. Number of 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)
4. What does it mean when we say that an algorithm X is asymptotically more efficient than Y?
Options:
- X will always be a better choice for small inputs
- X will always be a better choice for large inputs
- Y will always be a better choice for small inputs
- X will always be a better choice for all inputs
2. X will always be a better choice for large inputs
Explanation: In asymptotic analysis we consider growth of 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 following code:
int a = 0, i = N; while (i > 0) { a += i; i /= 2; } |
Options:
- O(N)
- O(Sqrt(N))
- O(N / 2)
- O(log N)
Output:
4. O(log N)
Explanation: We have to find the smallest x such that N / 2^x N
x = log(N)
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