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[HackerRank] Sorting : Mark and Toys (Python) 본문

Programming/코딩 1일 1문제

[HackerRank] Sorting : Mark and Toys (Python)

사용자 솜씨좋은장씨 2020. 3. 25. 15:29

Mark and Jane are very happy after having their first child. Their son loves toys, so Mark wants to buy some. There are a number of different toys lying in front of him, tagged with their prices. Mark has only a certain amount to spend, and he wants to maximize the number of toys he buys with this money.

Given a list of prices and an amount to spend, what is the maximum number of toys Mark can buy? For example, if prices = [ 1, 2, 3, 4 ]  and Mark has k = 7 to spend, he can buy items [ 1, 2, 3 ] for 6, or [ 3, 4 ] for 7 units of currency. He would choose the first group of 3 items.

 

Function Description

Complete the function maximumToys in the editor below. It should return an integer representing the maximum number of toys Mark can purchase.

maximumToys has the following parameter(s):

  • prices: an array of integers representing toy prices
  • k: an integer, Mark's budget

Input Format

The first line contains two integers, n and k, the number of priced toys and the amount Mark has to spend.
The next line contains n space-separated integers prices [ ]

 

Constraints

1 <= n <= 10^5

1 <= k <= 10^9
1 <= prices [ i  ] <= 10^9
A toy can't be bought multiple times.

 

Output Format

An integer that denotes the maximum number of toys Mark can buy for his son.

 

Sample Input

7 50
1 12 5 111 200 1000 10

Sample Output

4

Explanation

He can buy only 4 toys at most. These toys have the following prices: 1, 12, 5, 10.

 

Solution

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the maximumToys function below.
def maximumToys(prices, k):
    sort_prices = sorted(prices)

    count = 0

    for i in range(len(sort_prices)):
        k = k - sort_prices[i]

        if k < 0 :
            break
        count = count + 1

    return count

 

 

SOMJANG/CODINGTEST_PRACTICE

1일 1문제 since 2020.02.07. Contribute to SOMJANG/CODINGTEST_PRACTICE development by creating an account on GitHub.

github.com

 

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