Cracking the Coding Interview - Fooo

Solutions to Chapter 3 | Stacks and Queues
3 2 How would you design a stack which, in addition to push and pop, also has a function
min which returns the minimum element? Push, pop and min should all operate in
O(1) time
SOLUTION
1 1 3
Cracking the Coding Interview | Data Structures
pg 52
You can implement this by having each node in the stack keep track of the minimum beneath
itself Then, to find the min, you just look at what the top element thinks is the min
When you push an element onto the stack, the element is given the current minimum It sets
its “local min” to be the min
1 public class StackWithMin extends Stack {
2 public void push(int value) {
3 int newMin = Math.min(value, min());
4 super.push(new NodeWithMin(value, newMin));
5 }
6
7 public int min() {
8 if (this.isEmpty()) {
9 return Integer.MAX_VALUE;
10 } else {
11 return peek().min;
12 }
13 }
14 }
15
16 class NodeWithMin {
17 public int value;
18 public int min;
19 public NodeWithMin(int v, int min){
20 value = v;
21 this.min = min;
22 }
23 }
There’s just one issue with this: if we have a large stack, we waste a lot of space by keeping
track of the min for every single element Can we do better?
We can (maybe) do a bit better than this by using an additional stack which keeps track of
the mins
1 public class StackWithMin2 extends Stack {
2 Stack s2;
3 public StackWithMin2() {
4 s2 = new Stack();