Benchmark for http://stackoverflow.com/a/32924620/1977620

Main.java

import java.util.HashMap;
import java.util.Map;
import java.util.stream.Collectors;
import org.apache.commons.lang.RandomStringUtils;

public class Main {

    public static abstract class Test {
        abstract boolean start(String a, String b);

        long run(String a, String b) {
            long startTime = System.currentTimeMillis();
            start(a, b);
            long stopTime  = System.currentTimeMillis();
            return stopTime - startTime;
        }

    }

    public static Test test1 = new Test() {
        public boolean start(String a, String b) {
            Map<Character, Integer> countChars = new HashMap<>();
            // in Java 8. For older versions, it is also easy but more verbose
            for (char ch : a.toCharArray()) countChars.put(ch, countChars.getOrDefault(ch, 0) + 1);
            for (char ch : b.toCharArray()) {
                int count = countChars.getOrDefault(ch, 0);
                if (count < 1) return false;
                countChars.put(ch, count - 1);
            }
            return true;
        }
    };

    public static Test test2 = new Test() {
        public boolean start(String a, String b) {
            // Here we are counting occurrences of characters in the string
            Map<Integer, Long> aCounted = a.chars().boxed().collect(Collectors.groupingBy(o -> o, Collectors.counting()));
            Map<Integer, Long> bCounted = b.chars().boxed().collect(Collectors.groupingBy(o -> o, Collectors.counting()));

            // Now we're checking if the second string contains all the characters from the first
            return bCounted.keySet().stream().allMatch(
                    x -> bCounted.getOrDefault(x, 0l) >= aCounted.getOrDefault(x, 0l)
            );
        }
    };

    public static Test test3 = new Test() {
        public boolean start(String a, String b) {
            // Here we are counting occurrences of characters in the string
            Map<Integer, Long> aCounted = a.chars().parallel().boxed().collect(Collectors.groupingBy(o -> o, Collectors.counting()));
            Map<Integer, Long> bCounted = b.chars().parallel().boxed().collect(Collectors.groupingBy(o -> o, Collectors.counting()));

            // Now we're checking if the second string contains all the characters from the first
            return bCounted.keySet().parallelStream().allMatch(
                    x -> bCounted.getOrDefault(x, 0l) >= aCounted.getOrDefault(x, 0l)
            );
        }
    };

    public static void main(String[] args) {
        int[] size = {15000, 500000, 5000000, 50000000};
        Test[] tests = { test1, test2, test3 };

        for (int N : size) {
            String a = RandomStringUtils.randomAlphanumeric(N);
            String b = RandomStringUtils.randomAlphanumeric(N);

            // Warm-up JIT compiler
            for (Test test : tests) {
                test.start(a, b);
            }

            System.out.printf("N = %9d: ", N);
            for (Test test : tests) {
                long runTime = test.run(a, b);
                System.out.printf("%8dms", runTime);
            }

            System.out.println();

        }
    }
    
}