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interview prac/Heaps, Stacks, Queues/minimumOnStack/README.md

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+Note: Write a solution with O(operations.length) complexity, since this is what you would be asked to do during a real interview.
+
+Implement a modified stack that, in addition to using push and pop operations, allows you to find the current minimum element in the stack by using a min operation.
+
+Example
+
+For operations = ["push 10", "min", "push 5", "min", "push 8", "min", "pop", "min", "pop", "min"], the output should be
+minimumOnStack(operations) = [10, 5, 5, 5, 10].
+##### Stack과 Heap을 이용하여 문제를 풀었다.
+```java
+int[] minimumOnStack(String[] o) {
+ Queue<Integer> q = new PriorityQueue(3);
+ List<Integer> l = new ArrayList<>();
+ Stack<Integer> st = new Stack<>();
+ for(String s : o){
+ String t = s.substring(0,2);
+ int i=0;
+ if(t.equals("pu")){
+ i = Integer.valueOf(s.split(" ")[1]);
+ st.push(i);
+ q.add(i);
+ }else if(t.equals("mi")){
+ l.add(q.peek());
+ }else{
+ i = st.pop();
+ q.remove(i);
+ }
+ }
+ return l.stream().mapToInt(i->i).toArray();
+}
+```