import java.util.*;
public static void main(String[] args) {
    
    Scanner sc = new Scanner(System.in);
    System.out.print("Enter the size ??");
    int n = sc.nextInt();
    int[] marks = new int[n   1];
    
    for (int i = 0; i < n; i  ) {
        System.out.println("Enter "   i   " number ??");
        marks[i] = sc.nextInt();
    }
    
    System.out.println("The  following numbers are : ");
    for (int j = 0; j < marks.length - 1; j  ) {
        System.out.println(marks[j]   " ");
    }
    
    int max = marks[0];
    int min = marks[0];
    int s = marks.length;
    for (int i = 0; i < s; i  ) {
        if (marks[i] > max) {
            max = marks[i];
        }
        if (marks[i] < min) {
            min = marks[i];
        }
    }
    
    System.out.println("Max is "   max   " Min is "   min);
}

Output:

Enter the size ??2
Enter 0 number ??
56
Enter 1 number ??
56
The  following numbers are : 
56 
56 
Max is 56 Min is 0

CodePudding user response：

The size of mark array is 3 in the above example

 int[] marks = new int[n   1];

Let total number of marks is (n) : 2

We are creating the array with size of 3.

By default value for int primitives in java is 0

marks array is created with size 3

{56, 56, 0}

As per the logic the minimum value among these three elements are 0

CodePudding user response：

Hello and welcome to StackOverflow. Next time, febore you jump into the internet for help, please try this approach. It will solve your problem much quicker.

for (int i = 0; i < s; i  ) {
    System.out.println("DEBUG: number in index "   i   " is "   marks[i]);

    if (marks[i] > max) {
        System.out.println("DEBUG: number is greater than current max "   max);

        max = marks[i];
    }
    if (marks[i] < min) {
        System.out.println("DEBUG: number is smaller than current min "   min);

        min = marks[i];
    }
}

This process is called "debugging". It can be done by adding spurious amount of logging like above or by stepping through the code with a debugger.