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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
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.
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)
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;
}
}
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();
}
};
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)
}
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;
}
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