bideshbanerjee444's blog

Is this Code Ok to get saved from TLE and What will be the time Complexity of this???
By bideshbanerjee444, history, 2 years ago, In English

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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Comments Comments (8)
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2 years ago, # |
← Rev. 2   Vote: I like it +5 Vote: I do not like it

Complexity is O(nlogn), so it will pass.
Also, you can use custom invocation to check run time.

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    2 years ago, # ^ |
      Vote: I like it -33 Vote: I do not like it

    wrong is O(1)

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      2 years ago, # ^ |
        Vote: I like it +9 Vote: I do not like it

      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.

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        2 years ago, # ^ |
          Vote: I like it -36 Vote: I do not like it

        but n is constant (1e6)

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          2 years ago, # ^ |
            Vote: I like it +3 Vote: I do not like it

          So?

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          2 years ago, # ^ |
            Vote: I like it -8 Vote: I do not like it

          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 $$$O(1)$$$

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            2 years ago, # ^ |
              Vote: I like it +29 Vote: I do not like it

            he is trying to make joke about O(any constant) = O(1), and 10^6 log 10^6 is constant here

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              2 years ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              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.

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