3.5. sort stack exercise

03. Stacks and queues/3.5. sort_stack.md

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+# 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, `pop`ping them from s2 and `push`ing them in s1, and then `push` 5 to s2. Then `pop` 12 and 8 from s1 and `push` them to s2.
+
+```java
+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.