Is there any efficient way to do "shuffling" of vector - c++

Question

I have a large size unsorted array, each element contains a unique integer number,
std::vector<size_t> Vec= {1, 5, 3, 7, 18...}
I need to shuffle the vector in such a way, given a specific number, look for it and then swap it with the number in a new desired position. This swapping needs to be done many times.
Currently I use anther vector PositionLookup to remember&update the positions after every swapping. And I'm wondering is there any more efficient way/data structure that can help do this?
Current solution,
//look for a specific number "key" and swap it with the number in desired position "position_new"
void shuffle(key, position_new)
{
size_t temp = Vec[position_new]; // main vector
size_t position_old = PositionLookup[key]; // auxiliary vector
Vec[position_old] = temp;
PositionLookup[temp] = position_old;
Vec[position_new] = key;
PositionLookup[key] = position_new;
}

Answer 1

A couple microoptimizations to start with: If the vector has a fixed size, you could use a std::array or a plain C array instead of a std::vector. You can also use the most compact integer type that can hold all the values in the vector (e.g. std::int8_t/signed char for values in the interval [-128,127], std::uint16_t/unsigned short for values in the interval [0,65535], etc.)
The bigger optimization opportunity: Since the values themselves never change, only their indexes, you only need to keep track of the indexes.
Suppose for simplicity's sake the values are 0 through 4. In that case we can have an array
std::array<std::int8_t, 5> indices{{2, 3, 1, 4, 0}};
Which represents the index of its indices in an imaginary array, here 4, 2, 0, 1, 3. Or in other words indices[0] is 2, which is the index of 0 in the imaginary array.
Then to swap the positions of 0 and 1 you only need to do
std::swap(indices[0], indices[1]);
Which makes the indices array 3, 2, 1, 4, 0 and the imaginary array 4, 2, 1, 0, 3.
Of course the imaginary array's values might not be the same as its indices.
If the (sorted) values are something like -2, -1, 0, 1, 2 you could obtain the value from the index by adding 2, or if they're 0, 3, 6, 9, 12 you could divide by 3, or if they're -5, -3, -1, 1, 3 you could add 5 then divide by 2, etc.
If the values don't follow a defined pattern, you can create a second array to look up the value that goes with an index.
std::array<std::int8_t, 5> indices{{2, 3, 1, 4, 0}};
constexpr std::array<std::int8_t, 5> Values{{1, 3, 5, 7, 18}};
// Imaginary array before: 18, 5, 1, 3, 7
std::swap(indices[0], indices[1]);
// Imaginary array after: 18, 5, 3, 1, 7
const auto index_to_value = [&](decltype(indices)::value_type idx) noexcept {
return Values[idx];
};
const auto value_to_index = [&](decltype(Values)::value_type val) noexcept {
return std::lower_bound(Values.begin(), Values.end(), val)
- Values.begin();
};
It's the same thing if the values aren't known until runtime, just obviously the values lookup table can't be const or constexpr.
std::array<std::int8_t, 5> indices{{2, 3, 1, 4, 0}};
std::array<std::int8_t, 5> values; // Not known yet at compile-time
// ... set `values` at runtime to e.g. -93, -77, -64, 8, 56
// Imaginary array before: 56, -64, -93, -77, 8
std::swap(indices[0], indices[1]);
// Imaginary array after: 56, -64, -77, -93, 8
const auto index_to_value = [&](decltype(indices)::value_type idx) noexcept {
return values[idx];
};
const auto value_to_index = [&](decltype(values)::value_type val) noexcept {
return std::lower_bound(values.cbegin(), values.cend(), val)
- values.cbegin();
};