A simple solution i would think of is,

STEP-1: Create a array initially with size 10001 with all zeros.

STEP-2: We are going to decrement the value corresponding to the value that we read as input. for example if the first element of list A is 200, then array[200]--.. Similarly we decrement the values for list A.

STEP-3: Similarly when reading list B increment the values. Since the elements of list A are lost, the resulting values in the array after this operation will be positive.

STEP-4: Print all the index of array with value grater than 0. The numbers will be in asscending order aswell.

This solution can be optimized in a number of ways. Hint is "Xmax−Xmin<101".

Happy Coding!!

Java users note that if you're using HashMap to represent number frequencies, make sure you use the .equals() method to compare them and not == because Integer is an object while int is a primitive, that was causing me WA

in Python, if you know Counter from the standard collection library, you can do:

def missingNumbers(arr, brr):
    a = Counter(arr)
    b = Counter(brr)
    return sorted((b - a).keys())

Not sure why this question worth 45 points, its even much eaiser than some of those 15/20 points questions.

The difference between maximum and minimum number in the list B is less than or equal to 100 - this actually gave the clue away for solving this problem, without which the problem would have been much harder.

Simple in Python 3 :-)

from collections import Counter
n,arr=int(input()),list(map(int,input().split()))
m,brr=int(input()),list(map(int,input().split()))
a,b=Counter(arr),Counter(brr)
print(*(sorted((b-a).keys())))

Explanation:

Let us take an example :
n=10
arr=11 4 11 7 13 4 12 11 10 14
m=15
brr=11 4 11 7 3 7 10 13 4 8 12 11 10 14 12
a=Counter(arr)=Counter({11: 3, 4: 2, 7: 1, 13: 1, 12: 1, 10: 1, 14: 1})
b=Counter(brr)=Counter({11: 3, 4: 2, 7: 2, 10: 2, 12: 2, 3: 1, 13: 1, 8: 1, 14: 1})
b-a=Counter({7: 1, 3: 1, 10: 1, 8: 1, 12: 1})
(b-a).keys()=dict_keys([7, 3, 10, 8, 12])
sorted((b-a).keys())=3 7 8 10 12