In this HackerRank collection.counter() problem solution in python, we need to develop a python program that can read a list and then we need to print a dictionary that contains the number and number of occurrences on the output screen.
Problem solution in Python 2 programming.
from collections import Counter X = int(input()) sizes = Counter(map(int,raw_input().split())) N = int(input()) earnings = 0 for i in xrange(N): size,x = map(int,raw_input().split()) if sizes[size]>0: sizes[size]-=1 earnings += x print earnings
Problem solution in Python 3 programming.
# Enter your code here. Read input from STDIN. Print output to STDOUT import collections numShoes = int(input()) shoes = collections.Counter(map(int, input().split())) numCust = int(input()) income = 0 for i in range(numCust): size, price = map(int, input().split()) if shoes[size]: income += price shoes[size] -= 1 print(income)
Problem solution in pypy programming.
import collections numShoes = int(raw_input()) shoes = collections.Counter(map(int, raw_input().split())) numCust = int(raw_input()) money = 0 for i in range(numCust): size, price = map(int, raw_input().split()) if shoes[size]: money += price shoes[size] -= 1 print money
Problem solution in pypy3 programming.
# Enter your code here. Read input from STDIN. Print output to STDOUT from collections import Counter num_shoes = int(input()) sizes = [int(i) for i in input().split()] #2 3 4 5 6 8 7 6 5 18 num_cust = int(input()) d = Counter(sizes) earn = 0 for i in range(num_cust): #while(input() != null) req_i = [int(j) for j in input().split()] req_i_size = req_i[0] if req_i_size in d.keys(): if d[req_i_size] > 0: earn += req_i[1] d[req_i_size] -= 1 print(earn)
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