There is an array of integers. There are also disjoint sets, and , each containing integers. You like all the integers in set and dislike all the integers in set . Your initial happiness is . For each integer in the array, if , you add to your happiness. If , you add to your happiness. Otherwise, your happiness does not change. Output your final happiness at the end.
Note: Since and are sets, they have no repeated elements. However, the array might contain duplicate elements.
Constraints
Input Format
The first line contains integers and separated by a space. The second line contains integers, the elements of the array. The third and fourth lines contain integers, and , respectively.
Output Format
Output a single integer, your total happiness.
Sample Input
3 2 1 5 3 3 1 5 7
Sample Output
1
Explanation
You gain unit of happiness for elements and in set . You lose unit for in set . The element in set does not exist in the array so it is not included in the calculation.
Hence, the total happiness is .
Problem solution in Python 2 programming.
import sys
def main(): h = Happiness() input = get_input() m = input["m"] n = input["n"] n_array = input["n_array"] A_set = input["A_set"] B_set = input["B_set"] for num in n_array: if num in A_set: h.incr() elif num in B_set: h.decr() print h.val
# Enter your code here. Read input from STDIN. Print output to STDOUT n, m = input().split()
sc_ar = input().split()
A = set(input().split()) B = set(input().split()) print(sum([(i in A) - (i in B) for i in sc_ar]))
Problem solution in pypy programming.
# Enter your code here. Read input from STDIN. Print output to STDOUT n, m = raw_input().split()
arr = raw_input().split()
A = set(raw_input().split()) B = set(raw_input().split()) print sum([(i in A) - (i in B) for i in arr])
Problem solution in pypy3 programming.
# Enter your code here. Read input from STDIN. Print output to STDOUT numlst =input().split() l = input().split() A = set(input().split()) B = set(input().split()) print(len(list(filter(lambda x: x in A, l))) - len(list(filter(lambda x: x in B, l))))