How is make_heap in C++ implemented to have complexity of 3N?

Question

I wonder what\'s the algorithm of make_heap in in C++ such that the complexity is 3*N? Only way I can think of to make a heap by inserting elements have complexity of O(N Log N

Answer 1

You represent the heap as an array. The two elements below the i'th element are at positions 2*i+1 and 2*i+2. If the array has n elements then, starting from the end, take each element, and let it "fall" to the right place in the heap. This is O(n) to run.

Why? Well for n/2 of the elements there are no children. For n/4 there is a subtree of height 1. For n/8 there is a subtree of height 2. For n/16 a subtree of height 3. And so on. So we get the series n/22 + 2*n/23 + 3*n/24 + ... = (n/2)(1 * (1/2 + 1/4 + 1/8 + . ...) + (1/2) * (1/2 + 1/4 + 1/8 + . ...) + (1/4) * (1/2 + 1/4 + 1/8 + . ...) + ...) = (n/2) * (1 * 1 + (1/2) * 1 + (1/4) * 1 + ...) = (n/2) * 2 = n. So the total number of "see if I need to fall one more, and if so which way do I fall? comparisons comes to n. But you get round-off from discretization, so you always come out to less than n sets of swaps to figure out. Each of which requires at most 3 comparisons. (Compare root to each child to see if it needs to fall, then the children to each other if the root was larger than both children.)