ctci/03. Stacks and queues/3.5. sort_stack.md

3.5. Sort stack

Sort a stack such that the smallest items are on the top. You can use an additional temporary stack, but no arrays. Only push, pop, peek and isEmpty can be used

First idea

Example: [5, 3, 1, 6, 4]

I'm going to need another min variable in the data structure. The algorithm could go like this: Iterate stack until node.data == node.min, stack2.push(node.data). Then update the max, until second stack is full of nodes in right order and initial stack is empty. Then put the numbers back in initial stack.

This takes O(n) space and O(n^2) time.

Hints

• One way of sorting an array is to iterate through the array and insert each elem into a new array in sorted order, can you do this with a stack?
  • Yes, algorithm before
• Imagine stack2 is sorted, can you insert elemens into it in a sorted order? If you need extra storage, what could you use for extra storage?
  • I thought I couldn't modify functions. I can change push to iterate the stack and push the element in its correct position
• Keep stack2 sorted, with the biggest elements on top. Use stack1 for additional storage

So just iterate stack1 and add each element to stack2 in sorted mode, by having push modified. This takes O(nlogn)? time and O(n) space.

Solution

s1: [5,10,7] and s2: [12, 8, 3, 1]. We want to insert 5 into s2, sorted. To do that, we pop 5 from s1, hold it in a temporary variable, then move 12 and 8 to s1, popping them from s2 and pushing them in s1, and then push 5 to s2. Then pop 12 and 8 from s1 and push them to s2.

void sort(Stack<Integer> s) {
    Stack<Integer> r = new Stack<Integer>();
    while(!s.isEmpty()) {
        /* Insert each element in s in sorted order into r. */
        int tmp = s.pop();
        while(!r.isEmpty() && r.peek() > tmp) {
            s.push(r.pop());
        }
        r.push(tmp);
    }
    /* Copy the elements from r back into s. */
    while (!r.isEmpty()) {
        s.push(r.pop());
    }
 }

This is O(n^2) time and O(n) space. If we could use unlimited stacks, we could implement a modified quicksort or mergesort.

• Mergesort: create 2 extra stacks and divide stack1 into 2 parts, recursively sort each stack, and then merge them back together in sorted order into stack1. For each level of recursion, this means creating 2 additional stacks.
• Quicksort: create 2 extra stacks and divide the stack into the 2 stacks based on a pivot element. The 2 stacks would be recursively sorted, and then merged back together into stack1. This also involves creating 2 extra stacks per level of recursion.