Q. Count Subarrays(##)

There exists O(n) solution to count non-decreasing subarrays. find here - https://www.geeksforgeeks.org/count-strictly-increasing-subarrays/

/* ACCEPTED O(n^2) SOLUTION */
import java.util.*;

public class now {
    public static void Solution(int arr[]) {
        int n = arr.length;
        int count=n;    //initia;ise count by length becuase we consider the single elements as non-decreasing and count it here itself
        for (int i=0; i<n; i++) {   //first we are trying to find all the subarrays and then find the non decreasing sub arrays
            for(int j=i+1;j<arr.length;j++){
                if(arr[j]>=arr[j-1]) count++;
                else break;
            }
        }
        System.out.println(count);
    }

    public static void printArray(int arr[]) {
        for (int i = 0; i < arr.length; i++) {
            System.out.println(arr[i] + " ");
        }
        // System.out.println();
    }

    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int T;
        T = sc.nextInt();
        for (int t = 0; t < T; t++) {
            int size;
            size = sc.nextInt();

            int arr[] = new int[size];  //inputting array
            for (int i = 0; i < size; i++) {
                arr[i] = sc.nextInt();
            }

            Solution(arr);
        }
        sc.close();
    }
}


Link - https://codeforces.com/group/MWSDmqGsZm/contest/219774/problem/Q