There comes two methods from Collections class to find the elements with minimum and maximum values of a data structure. These methods can be applied to any Collection class (subclasses of Collection interface). In the following code ArrayList is used.
Following are the method signatures.
- static Object min(Collection col1): Returns the element with the minimum value in the Collection col1.
- static Object max(Collection col1): Returns the element with the maximum value in the Collection col1.
Example on Find Min Max values with min() max() meethods
import java.util.*;
public class CollectionsMaxMin
{
public static void main(String args[])
{
ArrayList firstList = new ArrayList();
firstList.add(50);
firstList.add(10);
firstList.add(20);
System.out.println("Elements firstList: " + firstList);
System.out.println("Minimum value element firstList: " + Collections.min(firstList));
System.out.println("Maximum value element firstList: " + Collections.max(firstList));
ArrayList secondList = new ArrayList();
secondList.add("five");
secondList.add("two");
secondList.add("six");
System.out.println("\nElements secondList: " + secondList);
System.out.println("Minimum value element secondList: " + Collections.min(secondList));
System.out.println("Maximum value element secondList: " + Collections.max(secondList));
}
}

Output screenshot on Find Min Max values with min() max()
ArrayList firstList = new ArrayList();
firstList.add(50); firstList.add(10); firstList.add(20);
An ArrayList object, firstList, is created with three integer elements 50, 10 and 20.
Collections.min(firstList);
Collections.max(firstList);
The min() and max() methods returns 10 and 50, the minimum and maximum values of firstList.
ArrayList secondList = new ArrayList();
secondList.add(“five”); secondList.add(“two”); secondList.add(“six”);
Another ArrayList object secondList is created with string elements "five", "two", and "six".
Collections.min(secondList);
Collections.max(secondList);
The above min() and max() methods with respect strings print five and two because f comes before t and s and last comes is t (after s). It is alphabetical order sorting.