rookiegang's blog

By rookiegang, history, 6 months ago, In English,

What is complexity of adding char in front of the string like s = c + s?

Because when I tested on c++ with code, its like O(1), but i think (but not sure then i ask here) i ever submitted at codeforces with this method, in O(n) loop then its get TLE (so the complexity O(n^2)), then i changed that part with other O(1) part then got AC.

What is complexity of adding on string like s = s2 + s1 with s1 is small string with length < 10 and s2 is n-long string?
What is complexity of adding on string like s = s1 + s2 with s1 is small string with length < 10 and s2 is n-long string?

Thank you codeforces community

 
 
 
 
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6 months ago, # |
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This has been answered by the STL.

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6 months ago, # |
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Adding a char in front of a string is always linear in length of the string.

In general, concatenating two strings will be linear in lengths of both strings.

However, if the first string is an rvalue, then the second one will be just appended to it. If appending doesn't cause the first string to reach its capacity, you can expect it to take time proportional to the length of the second string.

For example, the following code will take O(N^2) time:

    std::string a;
    for (int i = 0; i < N; ++i) {
      a = a + "xy";
    }

And the following code will take O(N) time:

    std::string a;
    for (int i = 0; i < N; ++i) {
      a = std::move(a) + "xy";
    }
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    6 months ago, # ^ |
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    wait , so for O(1) operation , I can add character in front of the string through second way( using move method ).

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      6 months ago, # ^ |
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      Unfortunately not. You can think of the second example as reusing memory of a and appending "xy" to it.

      You can think of adding a character in the front as s.insert(s.begin(), c), which is linear. Whether we reuse memory of a to create s or we make a copy won't change the complexity.

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    6 months ago, # ^ |
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    Also worth noting: a = a + "xy" is O(N) while a += "xy" is O(1) (amortized).