Next Permutation

/*
 * Problem : Process an array.
 * Input   : n, then n integers.
 * Output  : The next arrangement.
 * Approach: Find the pivot.
 * Time    : O(n)
 * Space   : O(1)
 * 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) {
        // Step 1: From the right, find the first place i where a[i] < a[i + 1]. The part after i is in descending order.
        int i = a.length - 2;
        while (i >= 0 && a[i] >= a[i + 1]) {
            i--;
        }

        // Step 2: If such a place exists, swap a[i] with the smallest value on its right that is bigger than a[i].
        if (i >= 0) {
            int j = a.length - 1;
            while (a[j] <= a[i]) {
                j--;
            }
            int temporary = a[i];
            a[i] = a[j];
            a[j] = temporary;
        }

        // Step 3: Reverse the part after i. It is now the smallest arrangement of those values.
        int left = i + 1;
        int right = a.length - 1;
        while (left < right) {
            int temporary = a[left];
            a[left] = a[right];
            a[right] = temporary;
            left++;
            right--;
        }
        return a;
    }

    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(" ")));
    }
}