How to work : sort function's third argument. [arr], []

Question

sort(arr.begin(), arr.end(), [arr](int& a, int& b) {}); // #1 - TLE
sort(arr.begin(), arr.end(), [](int& a, int& b) {}); // #2 - not TLE

What is the difference between [] and [arr] , and why does TLE #1 occur?

Answer 1

Assuming TLE is time limit exceeded for some online coding/competition site, [arr] will capture the collection as a copy so that it's accessible to the lambda (a comparator for the sorting logic).

For a big enough collection (or a small enough timeout), this could cause an issue.

You really don't need to capture the array since you shouldn't be doing anything with it. The contract between std::sort and the comparator is that you just indicate which of the two items is to be considered the larger, in a consistent manner.

I'm hoping that's not your complete code since you're meant to actually compare a and b somehow in the comparator function, and return the result of that comparison.

For example, a sorting the first few digits of π , largest to smallest, 9 5 4 3 1 1 :

#include <iostream>
#include <vector>
#include <algorithm>
using std::vector, std::sort, std::cout;

int main() {
    vector<int> vec { 3, 1, 4, 1, 5, 9 };
    sort(vec.begin(), vec.end(), [](int a, int b) {
        return b < a;
    });
    for (const auto &x: vec)
        cout << x << ' ';
    cout << '\n';
}

Note that we could have used std::greater<int>() rather than a lambda, but that wouldn't really show what you have to do in the lambda:-)