Asked  7 Months ago    Answers:  2   Viewed   41 times

I need to calculate the time complexity of the following code:

for (i = 1; i <= n; i++)
  for(j = 1; j <= i; j++)
   // Some code

Is it O(n^2)?



Yes, nested loops are one way to quickly get a big O notation.

Typically (but not always) one loop nested in another will cause O(n²).

Think about it, the inner loop is executed i times, for each value of i. The outer loop is executed n times.

thus you see a pattern of execution like this: 1 + 2 + 3 + 4 + ... + n times

Therefore, we can bound the number of code executions by saying it obviously executes more than n times (lower bound), but in terms of n how many times are we executing the code?

Well, mathematically we can say that it will execute no more than n² times, giving us a worst case scenario and therefore our Big-Oh bound of O(n²). (For more information on how we can mathematically say this look at the Power Series)

Big-Oh doesn't always measure exactly how much work is being done, but usually gives a reliable approximation of worst case scenario.

4 yrs later Edit: Because this post seems to get a fair amount of traffic. I want to more fully explain how we bound the execution to O(n²) using the power series

From the website: 1+2+3+4...+n = (n² + n)/2 = n²/2 + n/2. How, then are we turning this into O(n²)? What we're (basically) saying is that n² >= n²/2 + n/2. Is this true? Let's do some simple algebra.

  • Multiply both sides by 2 to get: 2n² >= n² + n?
  • Expand 2n² to get:n² + n² >= n² + n?
  • Subtract n² from both sides to get: n² >= n?

It should be clear that n² >= n (not strictly greater than, because of the case where n=0 or 1), assuming that n is always an integer.

Actual Big O complexity is slightly different than what I just said, but this is the gist of it. In actuality, Big O complexity asks if there is a constant we can apply to one function such that it's larger than the other, for sufficiently large input (See the wikipedia page)

Tuesday, June 1, 2021
answered 7 Months ago

The if condition will be true when j is a multiple of i; this happens i times as j goes from 0 to i * i, so the third for loop runs only i times. The overall complexity is O(n^4).

for (int i = 1; i < n; i++)
  for (int j = 0; j < i*i; j++)       // Runs O(n) times
      if (j % i == 0)                 // Runs O(n) × O(n^2) = O(n^3) times
          for (int k = 0; k < j; k++) // Runs O(n) × O(n) = O(n^2) times
              sum++;                  // Runs O(n^2) × O(n^2) = O(n^4) times
Friday, August 27, 2021
answered 4 Months ago
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