M.A.H.M.O.O.D's blog

Help in calculating time complixty
By M.A.H.M.O.O.D, history, 7 years ago, In English

Hi everyone.

I'm trying to solve this problem 559B - Equivalent Strings but I'm getting TLE here's my submission 26805388.

Now let's calculate the time complexity of this solution one call of function "is_equal" is O(n) in complexity because of the "substr" function which is linear in complexity.

Now for how many times we will call function "is_equal"...it should be O(log(n)) because each time we are halving the strings so the numbers of times I can do that is log(n) right ?

So the total time complexity should be O(NlogN) right ??? why am I getting TLE is there anything that I'm missing ??? I will be grateful for any help.

Thank you for reading.

:)

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7 years ago, # |
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You call your function 4 times each recursive call in worst case. What makes you think it's logN complexity?

Assuming each time your N gets halved this gives you logN depth but the number of sub problems is 4 to the power of something. ( Level i has 4^i subproblems). I think it's summation of powers of 4 from 0 till logN, something along these lines.

In binary search you only branch to one sub problem, here you branch to 4 instead so you can't use the same reasoning.

Hope that helps!

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7 years ago, # |
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Nope, complexity is O(n2) in your case. Your solution makes much more than calls (at the very least it makes O(n) calls because is_equal can be called from any substring of length 1).

If is_equal is called from two substrings of length n each, then it makes approximately n operations by itself and calls is_equal recursively four times, passing strings of length n/2 each time. If we denote total running time of is_equal as T(n) (here n is length of the arguments), then we get the following recurrence: . It can be solved using master theorem: T(n) = O(n2).

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