The time complexity of the given code is O(N^2).

Here's why:

1. The outer loop runs N times, where N is the input size.
2. For each iteration of the outer loop, the inner loop runs N - i times, where i is the current iteration of the outer loop.
3. The number of operations that the inner loop performs is proportional to the number of times it runs. Therefore, the total number of operations is the sum of the series 1 + 2 + 3 + ... + N, which is (N*(N+1))/2.
4. This sum is proportional to N^2, so the time complexity of the code is O(N^2).

Question

The time complexity of the given code is O(N^2).

Here's why:

1. The outer loop runs N times, where N is the input size.
2. For each iteration of the outer loop, the inner loop runs N - i times, where i is the current iteration of the outer loop.
3. The number of operations that the inner loop performs is proportional to the number of times it runs. Therefore, the total number of operations is the sum of the series 1 + 2 + 3 + ... + N, which is (N*(N+1))/2.
4. This sum is proportional to N^2, so the time complexity of the code is O(N^2).

Knowee AI · Accepted Answer

The time complexity of the given code is O(N^2).

Here's why:

1. The outer loop runs N times, where N is the input size.
2. For each iteration of the outer loop, the inner loop runs N - i times, where i is the current iteration of the outer loop.
3. The number of operations that the inner loop performs is proportional to the number of times it runs. Therefore, the total number of operations is the sum of the series 1 + 2 + 3 + ... + N, which is (N*(N+1))/2.
4. This sum is proportional to N^2, so the time complexity of the code is O(N^2).

What is the time complexity of the following code : int a = 0; for (i = 0; i < N; i++) { for (j = N; j > i; j--) { a = a + i + j; } }

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