# How to prove that the time complexity of this algorithm is O($\sqrt{N}$)?

  int n;
cin >> n;
int sum = 0;
for (int i = 1; sum <= n; i++) {
sum += i;
}


If I assumed that $$N = 100$$, the loop will run $$13$$ steps, which is almost the square root of $$N$$, if $$N = 10000$$, the loop will run $$141$$ steps, which is almost the square root of $$N$$, but I don't know how to prove that, I only know it by intuition

The loop iterates $$k$$ times, where $$k$$ is minimal such that $$\sum_{i=1}^k i>n$$. We know that $$\sum_{i=1}^k i = \tfrac12k(k+1)$$, so you just need to solve for $$k$$.