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Home/ Questions/Q 910769
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T17:05:40+00:00 2026-05-15T17:05:40+00:00

So I was looking at this code from a textbook: for (int i=0; i<N;

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So I was looking at this code from a textbook:

for (int i=0; i<N; i++)
   for(int j=i+1; j<N; j++)

The author stated that the inner for-loop iterates for exactly N*(N-1)/2 times but gives no basis for how he arrived to such an equation. I understand N*(N-1) but why divide by 2? I ran the code myself and sure enough when N is 10, the inner loop iterates 45 times (10*9/2).

I messed around with the code myself and tried the following (assigned only i to j):

for (int i=0; i<N; i++)
   for(int j=i; j<N; j++)

With N = 10, this results in 55. So I’m having trouble understanding the underlying math here. Sure I could just plug in all the values and bruteforce my way through the problem, but I feel there is something essential and very simple I’m missing. How would you come up with an equation for describing the for loop I just constructed? Is there a way to do it without relying on the outputs? Would really appreciate any help thanks!

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  1. Editorial Team
    Editorial Team
    2026-05-15T17:05:41+00:00Added an answer on May 15, 2026 at 5:05 pm

    Think about what happens each time the outer loop iterates. The first time, i == 0, so the inner loop starts at 1 and runs to N-1, which is N-1 iterations in total. The next time through the outer loop, i has incremented to 1, so the inner loop starts at 2 and runs up to N-1, for a total of N-2 iterations. And that pattern continues: the third time through the outer loop, you get N-3 iterations, the fourth time through, N-4, etc. When you get to the last iteration of the outer loop, i == N-1, so the inner loop starts with j = N and stops immediately. So that’s zero iterations.

    The total number of iterations is the sum of all these numbers:

    (N-1) + (N-2) + (N-3) + ... + 1 + 0
    

    To look at it another way, this is just the sum of the positive integers from 1 to N-1. The result of this sum is called the (N-1)th triangular number, and Wikipedia explains how you can find that the formula for the n’th triangular number is n(n+1)/2. But here you have the (N-1)th triangular number, so if you set n=N-1, you get

    (N-1)(N-1+1)/2 = N(N-1)/2
    
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