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push_back() simplification Your push_back() function seems overly complicated. This part: if (m_tail != nullptr) { auto at = iterator(m_tail); auto node = new Node; node->next = at.m …

Reducing number of writes Right now, you swap neighboring elements until the newest element reaches its proper place. This takes 2n array writes to move an element n spots. Instead of swapping, you …

Bug Your formula for the surface area of a Box is wrong: double surfaceArea() const override { return (2*length*height) + (4*width*height); } It should be: double surfaceArea() const override { ret …

Remove I find it interesting that you have a special case for removing the root where if it has one child, you reroot the tree on the child. That is actually a good move because it always reduces th …

More efficient way Your current algorithm runs in \$O(N^2)\$ time. At the time I started thinking about this, I did not read Jerry Coffin's link that demonstrated an \$O(N)\$ solution, so I came up …

ABA problem I was able to break your queue (but it wasn't easy). I inserted some code to freeze one thread here in consume(): do{ // The divider's next pointer points to the next node wit …

Too complex With the problem constrained to: The elements of the sequence are integers in range [0, 99] The length of the sequence is in range [0, 50] the program should be very simple. Instead, yo …

Speeding things up There are several things you could do to speed things up: Instead of looping from 100..999, go from 999..100 instead. That way, if you find a palindrome, you can break from the j …

Faster sorting If the only purpose of the distance computation is for comparison, then you can speed it up by removing the call to sqrt(). You'll end up comparing the squares of the distances, which …

Bug This line in your quicksort isn't right: int pivot = toSort[(rand() % endPtr - beginPtr + 1) + beginPtr]; It should be: int pivot = toSort[rand() % (endPtr - beginPtr) + beginPtr]; Notice that …

Pop() has problem Consider what happens if your queue is in the following state with two jobs: m_top = 0 m_bottom = 2 The owner thread calls pop(), and loads in to local variables: top = 0 bott …

Stop printing so much I ran your program and found that it spent most of its time printing every single combination (even though none of them were solutions). After I removed the line that prints the …

Bug When you pop an element off, you increment iHead before reading the return element from elements. If that thread pauses between the increment and the read, another thread can push a new element i …

More clear by using extra argument? If you passed in a count by reference to each function, and each function incremented this count instead of returning a new count in a tuple, the code might look cl …

Not an O(N^2) algorithm You create a vector that contains every (i,j) pair. This vector contains O(N^2) elements. For each pair, you loop across the original vector of N elements. Outer loop = O( …