for(int i = 2; i <= int(1e6); i++){
for(int j = 0; j <= int(1e6); j += i){
//Here anything which takes O(1) time.....
} }
for(int i = 2; i <= int(1e6); i++){
for(int j = 0; j <= int(1e6); j += i){
//Here anything which takes O(1) time.....
} }
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Complexity is O(nlogn), so it will pass.
Also, you can use custom invocation to check run time.
wrong is O(1)
Notice that
j += i. So the overall complexity is $$$O(\frac{n}{1} + \frac{n}{2} + \dots + \frac{n}{n}) = O(n \log n)$$$I dont think computer are smart enough to solve that kind of loop.
but n is constant (1e6)
So?
It is more complex then you think just a $$$O(1)$$$ and compiler can predict all 2-loops.
Some problem are even already hard to make $$$O(n \log n) \rightarrow O(n \log \log n)$$$ even if the inside function run in $$$O(1)$$$
he is trying to make joke about O(any constant) = O(1), and 10^6 log 10^6 is constant here
Ok I get the joke. And well it is also true as there is the fact that the loop $$$10^6$$$ parts are not dependent on a variable taken from the input.