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comments difficulty edit_url rating source tags
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Easy
1248
Biweekly Contest 9 Q1
Greedy
Array
Sorting

中文文档

Description

You have some apples and a basket that can carry up to 5000 units of weight.

Given an integer array weight where weight[i] is the weight of the ith apple, return the maximum number of apples you can put in the basket.

 

Example 1:

Input: weight = [100,200,150,1000]
Output: 4
Explanation: All 4 apples can be carried by the basket since their sum of weights is 1450.

Example 2:

Input: weight = [900,950,800,1000,700,800]
Output: 5
Explanation: The sum of weights of the 6 apples exceeds 5000 so we choose any 5 of them.

 

Constraints:

  • 1 <= weight.length <= 103
  • 1 <= weight[i] <= 103

Solutions

Solution 1: Greedy Algorithm

To maximize the number of apples, we should try to minimize the weight of the apples. Therefore, we can sort the weights of the apples, and then put them into the basket in ascending order until the weight of the basket exceeds $5000$. We then return the number of apples in the basket at this point.

If all the apples can be put into the basket, then we return the total number of apples.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the number of apples.

Python3

class Solution:
    def maxNumberOfApples(self, weight: List[int]) -> int:
        weight.sort()
        s = 0
        for i, x in enumerate(weight):
            s += x
            if s > 5000:
                return i
        return len(weight)

Java

class Solution {
    public int maxNumberOfApples(int[] weight) {
        Arrays.sort(weight);
        int s = 0;
        for (int i = 0; i < weight.length; ++i) {
            s += weight[i];
            if (s > 5000) {
                return i;
            }
        }
        return weight.length;
    }
}

C++

class Solution {
public:
    int maxNumberOfApples(vector<int>& weight) {
        sort(weight.begin(), weight.end());
        int s = 0;
        for (int i = 0; i < weight.size(); ++i) {
            s += weight[i];
            if (s > 5000) {
                return i;
            }
        }
        return weight.size();
    }
};

Go

func maxNumberOfApples(weight []int) int {
	sort.Ints(weight)
	s := 0
	for i, x := range weight {
		s += x
		if s > 5000 {
			return i
		}
	}
	return len(weight)
}

TypeScript

function maxNumberOfApples(weight: number[]): number {
    weight.sort((a, b) => a - b);
    let s = 0;
    for (let i = 0; i < weight.length; ++i) {
        s += weight[i];
        if (s > 5000) {
            return i;
        }
    }
    return weight.length;
}