Below is how to implement the functions if Math.min() and Math.max() did not exist.

Functions have an arguments object, which you can iterate through to get its values.

It's important to note that Math.min() with no arguments returns Infinity, and Math.max() with no arguments returns -Infinity.

function min() {
  var result= Infinity;
  for(var i in arguments) {
    if(arguments[i] < result) {
      result = arguments[i];
    }
  }
  return result;
}

function max() {
  var result= -Infinity;
  for(var i in arguments) {
    if(arguments[i] > result) {
      result = arguments[i];
    }
  }
  return result;
}

//Tests
console.log(min(5,3,-2,4,14));       //-2
console.log(Math.min(5,3,-2,4,14));  //-2

console.log(max(5,3,-2,4,14));       //14
console.log(Math.max(5,3,-2,4,14));  //14

console.log(min());                  //Infinity
console.log(Math.min());             //Infinity

console.log(max());                  //-Infinity
console.log(Math.max());             //-Infinity

In this case, I can also think of $D$ as a function in $x$:

$D(x) = \max(0, M(x)) = \begin{cases} M(x) &\text{if}\ M(x) >0, \\ 0 &\text{otherwise}.\end{cases}$

The result is essentially $M$ "cut off" at $0$.

You can simply used in-build java Collection and Arrays to sort out this problem. You just need to import them and use it.

Please check below code.

import java.util.Arrays;
import java.util.Collections;

public class getMinNMax {
    public static void main(String[] args) {

        Integer[] num = { 2, 11, 55, 99 };

        int min = Collections.min(Arrays.asList(num));
        int max = Collections.max(Arrays.asList(num));

        System.out.println("Minimum number of array is : " + min);
        System.out.println("Maximum number of array is : " + max);
    }
}