README_EN.md

1833. Maximum Ice Cream Bars

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Description

It is a sweltering summer day, and a boy wants to buy some ice cream bars.

At the store, there are n ice cream bars. You are given an array costs of length n, where costs[i] is the price of the ith ice cream bar in coins. The boy initially has coins coins to spend, and he wants to buy as many ice cream bars as possible. 

Note: The boy can buy the ice cream bars in any order.

Return the maximum number of ice cream bars the boy can buy with coins coins.

You must solve the problem by counting sort.

 

Example 1:

Input: costs = [1,3,2,4,1], coins = 7
Output: 4
Explanation: The boy can buy ice cream bars at indices 0,1,2,4 for a total price of 1 + 3 + 2 + 1 = 7.

Example 2:

Input: costs = [10,6,8,7,7,8], coins = 5
Output: 0
Explanation: The boy cannot afford any of the ice cream bars.

Example 3:

Input: costs = [1,6,3,1,2,5], coins = 20
Output: 6
Explanation: The boy can buy all the ice cream bars for a total price of 1 + 6 + 3 + 1 + 2 + 5 = 18.

 

Constraints:

• costs.length == n
• 1 <= n <= 105
• 1 <= costs[i] <= 105
• 1 <= coins <= 108

Solutions

Pay attention to the data range. The question can easily mislead us to use the 01 backpack (it will overtime). In fact, this question is a simple "greedy problem" (choose low-priced ice cream first)

Python3

class Solution:
    def maxIceCream(self, costs: List[int], coins: int) -> int:
        costs.sort()
        for i, c in enumerate(costs):
            if coins < c:
                return i
            coins -= c
        return len(costs)

Java

class Solution {
    public int maxIceCream(int[] costs, int coins) {
        Arrays.sort(costs);
        int n = costs.length;
        for (int i = 0; i < n; ++i) {
            if (coins < costs[i]) {
                return i;
            }
            coins -= costs[i];
        }
        return n;
    }
}

C++

class Solution {
public:
    int maxIceCream(vector<int>& costs, int coins) {
        sort(costs.begin(), costs.end());
        int n = costs.size();
        for (int i = 0; i < n; ++i) {
            if (coins < costs[i]) return i;
            coins -= costs[i];
        }
        return n;
    }
};

Go

func maxIceCream(costs []int, coins int) int {
	sort.Ints(costs)
	for i, c := range costs {
		if coins < c {
			return i
		}
		coins -= c
	}
	return len(costs)
}

JavaScript

/**
 * @param {number[]} costs
 * @param {number} coins
 * @return {number}
 */
var maxIceCream = function (costs, coins) {
    costs.sort((a, b) => a - b);
    const n = costs.length;
    for (let i = 0; i < n; ++i) {
        if (coins < costs[i]) {
            return i;
        }
        coins -= costs[i];
    }
    return n;
};

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