Arrangement of arrays such that maximum number of elements are greater in one array

user_9879525 0
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Assuming I have two arrays A and B (Both with equal number of elements)

int A[] = {40,50,70};
int B[] = {80,60,45};

I have to rearrange array A in such a way that maximum number of elements in Array A are greater than their respective elements in array B.

In this case, rearranging A as {40,70,50} would yield the required result.

What would be the most optimal way of going this?

arrays algorithm c++
2 Answers

I would use something like:

std::vector<int> f(std::vector<int> A, const std::vector<int>& B)
{
    std::vector<std::size_t> indexes(B.size());
    std::iota(indexes.begin(), indexes.end(), 0);

    std::sort(A.begin(), A.end(), std::greater<>{});
    std::sort(indexes.begin(), indexes.end(),
              [&B](std::size_t lhs, std::size_t rhs){ return B[lhs] > B[rhs]; });

    auto it = A.begin();
    auto rit = A.rbegin();
    std::vector<int> res(A.size());
    for (auto index : indexes) {
        if (*it > B[index]) {
            res[index] = *it;
            ++it;
        } else {
            res[index] = *rit;
            ++rit;
        }
    }
    return res;
}

Demo

Complexity: O(n log n).

user_26845390

Assume that we can also rearrange B for now. Clearly, it's always optimal to match the largest elements in A to lesser elements in B. Therefore, we can iterate over elements in A in decreasing order and try to pair them with lesser elements in B.

For each element in A, we'll want to use the greatest unused element in B which is less than the element in A, since we want to save the smaller elements in B for use later. We can sort B and find this element using binary search or by maintaining a pointer to the greatest unused element. If at any point we can't find an unused element in B that is less than the next element in A, we know that we won't be able to match the rest of the elements in A to lesser elements in B so we can match the remaining elements arbitrarily. Now we can rearrange the pairs so that the elements in B will be in their original order and we have a solution.

user_92545390
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