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We don't count the moment that the condition of the loop fails, since the sequence of statements is not going to be executed at all ;)

As for the English, make sure you keep up with this forum......

Let's solve it :)
Σ[base:i=0, upperBound:i=N-1] Σ[base:j = i+1, upperBound:j = N-1] 1

= Σ[base:i=0, upperBound:i=N-1] (N - 1 - i - 1 + 1) 1

= Σ[base:i=0, upperBound:i=N-1] ( N - 1 - i )

=...

The second one is true because of linearity...

The first one:

j will take the value of i+1... As a result, it will start from value 1, when you meet the inner loop for first time... Then, when...

I thought you implied that this way of declaring a 2D array was incorrect.

You are welcome :)

PS - Χρόνια πολλά για την Εθνική μας εορτή!

Well, this goes out of bounds. But let's say we have i and j (the complexity won't change). When I say complexity, I mean time complexity.

The complexity will be Ο(n²).
Τhe explanation is as you...