Kotlin coding test course, creating an ascending sequence with a stack

Problem Description

Given an integer n and a sequence of n integers, your task is to implement a method to sort the given sequence
in ascending order using a stack. You need to process the sequence taking advantage of the stack’s LIFO (Last In, First Out) property,
and the maximum number of elements that can enter the stack is n.

Input Format

• The first line contains an integer n (1 ≤ n ≤ 100,000).
• The second line contains n integers separated by spaces (1 ≤ each number ≤ 1,000,000), and the sequence may contain duplicates.

Output Format

Output the series of stack operations (push, pop) used to sort the sequence in ascending order.
If it is not possible to sort the sequence using the stack, you should output “NO”.

Example Input

    5
    4 3 6 8 5
    

Example Output

    PUSH 4
    PUSH 3
    POP 3
    PUSH 6
    POP 6
    PUSH 5
    POP 5
    POP 4
    

Problem Solution

This problem requires using a stack to sort the input sequence in ascending order.
Let’s approach the problem with the following steps.

Algorithm Design

1. **Input Handling**: Read the integer n and the sequence.
2. **Initialize Stack and Output List**: Initialize the list to store the sequence and the stack.
3. **Initialize Target Value**: Set the next target value that should be in order as 1.
4. **Traverse the Sequence**: Loop through the input sequence,
– If the current number is equal to the target value, perform the pop operation.
– If it is different from the target value, push it onto the stack.
5. **Empty the Stack**: If there are values left in the stack, pop and output them.
6. **Output**: Finally, display the sorted sequence or output “NO”.

Coding Implementation

If the above algorithm is implemented in Kotlin, it looks like this:

fun main() {
    val n = readLine()!!.toInt()
    val sequence = readLine()!!.split(" ").map { it.toInt() }
    val stack = mutableListOf()
    val output = mutableListOf()
    var current = 1
    var index = 0

    while (current <= n) {
        while (index < n && (stack.isEmpty() || stack.last() != current)) {
            stack.add(sequence[index++])
            output.add("PUSH ${stack.last()}")
        }
        if (stack.isNotEmpty() && stack.last() == current) {
            stack.removeAt(stack.size - 1)
            output.add("POP $current")
        } else {
            println("NO")
            return
        }
        current++
    }

    output.forEach { println(it) }
}

Key Considerations

There are several important considerations in solving this problem. First, you need to understand the concept of the stack
and implement it in code, utilizing the stack effectively in certain situations. Additionally,
since the input numbers must be sorted in ascending order, a logic that guarantees this is needed.
The maximum size of the stack is n, allowing for an O(n) space complexity.

Conclusion

In this lecture, we addressed the problem of creating an ascending sequence using a stack. We could confirm
our understanding of basic stack operations and their utilization through actual code implementation.
The stack data structure is covered in many algorithm problems, so I hope you learn how to utilize stacks through this problem.