Next Greater Element

/*
 * Problem : Process an array.
 * Input   : n, then n integers.
 * Output  : The next greater values.
 * Approach: A monotonic stack.
 * Time    : O(n)
 * Space   : O(n)
 * Java    : 8 or newer.
 */
import java.util.Scanner;
import java.util.Arrays;
import java.util.stream.Collectors;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static int[] solve(int[] a) {
        int[] result = new int[a.length];
        java.util.Arrays.fill(result, -1);

        // Step 1: The stack keeps the indexes of elements that still wait for a greater element.
        java.util.ArrayDeque<Integer> stack = new java.util.ArrayDeque<>();

        for (int i = 0; i < a.length; i++) {
            // Step 2: The current element is greater than the waiting ones on top. It is their answer.
            while (!stack.isEmpty() && a[stack.peek()] < a[i]) {
                int waiting = stack.pop();
                result[waiting] = a[i];
            }

            // Step 3: The current element now waits for its own answer.
            stack.push(i);
        }
        return result;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int size = scanner.nextInt();
        int[] a = new int[size];
        for (int i = 0; i < size; i++) {
            a[i] = scanner.nextInt();
        }
        System.out.println(Arrays.stream(solve(a)).mapToObj(String::valueOf).collect(Collectors.joining(" ")));
    }
}