Form Biggest Number [codingblocks]

You are provided an array of numbers. You need to arrange them in a way that yields the largest value.

Input Format
First line contains integer t which is number of test case.
For each test case, you are given a single integer n in the first line which is the size of array A[] and next line contains n space separated integers denoting the elements of the array A .

Constraints
1<=t<=100
1<=m<=100
1<=A[i]<=10^5

Output Format
Print the largest value.

Sample Input
1
4
54 546 548 60

Sample Output
6054854654

Explanation
Upon rearranging the elements of the array , 6054854654 is the largest possible number that can be generated.

Java Code:

IDE Code: https://ide.geeksforgeeks.org/jwhvpJ4eNe

/*
Amit Kumar
12-12-2020
IDE Code: https://ide.geeksforgeeks.org/jwhvpJ4eNe
*/


package String;
import java.util.Scanner;

public class String_PiyushAndMagicalPark {
    public static Scanner scanner = new Scanner(System.in);
    public static void main(String[] args) {
        int row = scanner.nextInt();
        int col = scanner.nextInt();
        int minStrength = scanner.nextInt();
        int currentStrength = scanner.nextInt();

        String[][] park = new String[row][col];
        for (int i=0; i<row; i++) {
            for (int j=0; j<col; j++) {
                park[i][j] = scanner.next();
            }
        }

        int finalStrength = getPiyushAndMagicalParkFinalStrength(park, row, col, minStrength, currentStrength);
        if (finalStrength > minStrength) {
            System.out.println("Yes");
            System.out.println(finalStrength);
        } else {
            System.out.println("No");
        }
    }

    private static int getPiyushAndMagicalParkFinalStrength(String[][] park, int row, int col, int minStrength, int currentStrength) {
        for (int i=0; i<row; i++) {
            for (int j=0; j<col; j++) {
                String currValue = park[i][j];

                if (currentStrength < minStrength) {
                    return currentStrength;
                }

                switch (currValue) {
                    case "." : currentStrength = currentStrength - 2; break;
                    case "*" : currentStrength = currentStrength + 5; break;
                    case "#" : j = col - 1; break;
                }

                if (j != col - 1) {
                    currentStrength--;
                }
            }
        }
        return currentStrength;
    }
}