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Time Complexity of '+' with strings Giving TLE??
By Focus_2000, history, 5 years ago, In English

Problem: A. Cut and Paste when i am using

s=s+t; [this is giving me TLE on 37 test ] [128332169](https://codeforces.me/contest/1280/submission/128332169)

and s+=t; is Accepted 128332405 20 ms

can anyone tell me why such happen.???

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5 years ago, hide # |
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Vote: I like it +37 Vote: I do not like it

s = s + t; this line will create new string object with value s + t and assign it to s, complexity $$$O(|s| + |t|)$$$

s += t; this line just simply append t to s, complexity $$$O(|t|)$$$

#include <bits/stdc++.h>
using namespace std;

signed main(){
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

    string s = "";

    time_t start = clock();
    for(int i = 0; i < 1000; i++) s = s + "a";
    cout << clock() - start << '\n';

    start = clock();
    for(int i = 0; i < 1000; i++) s += "a";
    cout << clock() - start;
}

This code produced the following output on my system

100
4

So avoid using assignment operator!

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    5 years ago, hide # ^ |
    ← Rev. 3  
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    Thanks. Found where i am wrong.

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

    When I'm running your code on Dev C++ it is printing 0 0, but in the online C++ compiler, it is giving the right answer. Can you please tell me what is wrong with the Dev C++ compiler? Thank You:)

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      5 years ago, hide # ^ |
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      I guess CLOCKS_PER_SEC on your system has the value 1000. So just add consuming stuff between 2 clock() call, you'll see the difference:

      time_t start = clock();
BogoSort();
cout << clock() - start;

      For better precision, I suggest using chrono. IMO using clock() is better as it's really easy to remember the syntax.

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5 years ago, hide # |
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I think the '+' operator's complexity is O(2*(n1 + n2)), where n1 and n2 are the lengths of the strings s and t.
The complexity of the operator '+=' might only be O(2*n2) in some implementations on the other hand as only the append() operation is to be performed.
A Stack Overflow Thread
If there is a huge difference between the lengths of s and t, this could effectively work out as the difference between an O(N) algorithm and an O(1) algorithm, and could possibly lead to TLE.

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