Sorting a vector in descending order within two ranges

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Say I have a vector of integers:

std::vector<int> indices;
for (int i=0; i<15; i++) indices.push_back(i);

Then I sort it in descending order:

sort(indices.begin(), indices.end(), [](int first, int second) -> bool{return indices[first] > indices[second];})
for (int i=0; i<15; i++) printf("%i\n", indices[i]);

This produces the following:

14
13
12
11
10
9
8
7
6
5
4
3
2
1
0

Now I want to have the numbers 3, 4, 5, and 6 to be moved to the end, and keep the descending order for them (preferably without having to use sort for the second time). I.e., here is what I want:

14
13
12
11
10
9
8
7
2
1
0
6
5
4
3

How should I modify the comparison function of the std::sort to achieve that?

stdvector vector c++11 sorting c++
3 Answers

Your comparison function is wrong since the values you get as first and second are the elements of the std::vector. Therefore, there is no need to use them as indices. So, you need to change

return indices[first] > indices[second];

to

return first > second;

Now, regarding the problem you try to solve...

You can leave 3, 4, 5 and 6 out of comparison with other elements and still compare it with each other:

std::sort(
    indices.begin(), indices.end(),
    [](int first, int second) -> bool {
        bool first_special = first >= 3 && first <= 6;
        bool second_special = second >= 3 && second <= 6;
        if (first_special != second_special)
            return second_special;
        else
            return first > second;
    }
);

Demo

user_55173780

Functions from the standard algorithms library like iota, sort, find, rotate and copy would make your life easier. Your example comes down to:

#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
#include <iterator>


int main()
{
  std::vector<int> indices(15);
  std::iota(indices.begin(), indices.end(), 0);
  std::sort(indices.begin(), indices.end(), std::greater<>());

  auto a = std::find(indices.begin(), indices.end(), 6);
  auto b = std::find(indices.begin(), indices.end(), 3);
  std::rotate(a, b + 1, indices.end());

  std::copy(indices.begin(), indices.end(), std::ostream_iterator<int>(std::cout, "\n"));
  return 0;
}

Output:

14
13
12
11
10
9
8
7
2
1
0
6
5
4
3

@TedLyngmo in the comments makes the good point that it could/should be improved with:

auto a = std::lower_bound(indices.begin(), indices.end(), 6, std::greater<int>{});
auto b = a + 4;
user_13320410

Solution 1

Straightforward approach with a non-linear comparator.

inline constexpr bool SpecialNumber(const int n) noexcept {
  return n < 7 && 2 < n;
}

void StrangeSortSol1(std::vector<int>* v) {
  std::sort(v->begin(), v->end(), [](const int a, const int b) noexcept {
    const bool aSpecial = SpecialNumber(a);
    const bool bSpecial = SpecialNumber(b);

    if (aSpecial && bSpecial) return b < a;
    if (aSpecial) return false;
    if (bSpecial) return true;
    return b < a;
  });
}

Solution 2

Using std::algorithms (partition)!

inline constexpr bool SpecialNumber(const int n) noexcept {
  return n < 7 && 2 < n;
}

void StrangeSortSol2(std::vector<int>* v) {
  auto pivot = std::partition(v->begin(), v->end(), std::not_fn(SpecialNumber));
  std::sort(v->begin(), pivot, std::greater{});
  std::sort(pivot, v->end(), std::greater{});
}

Performance Considerations

It may look like the second solution is slower because of the overhead of partition. Probably it's not, because of cache and miss-branch prediction in modern processors.

Benchmark

user_60401810
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