2024-09-04: In go, given an array happiness of length n, representing the happiness value of each child, and a positive integer k, we need to select k children from these n children.
During the screening process, each time a child is selected in a round, the Happiness value of all children not yet selected is reduced by 1. Note that the Happiness value cannot be reduced to a negative number; it can only be reduced if it is a positive number.
Our goal is to maximize the sum of the happiness values of the selected k children as much as possible.
Input: happiness = [1,2,3], k = 2.
Output: 4.
Explanation: Choose 2 children as follows:
1. Select the child with a happiness value of 3. The happiness value of the remaining children becomes [0,1].
2. Select the child with a happiness value of 1. The happiness value of the remaining children becomes [0]. Note that the happiness value cannot be less than 0.
The sum of the selected children's happiness values is 3 + 1 = 4 .
Answer 2024-09-04:
chatgpt
Title from leetcode3075.
The broad steps are as follows:
1. Sort the child's array of happiness values happiness in descending order.
2. Select the first k children with the highest happiness values from the sorted array. The sum of the happiness values of these children is the desired value.
3. Among the selected k children, determine, child by child, whether the happiness value is greater than or equal to the index value of the current location, and if so, subtract the happiness value from the current index value and add it to the final result, which represents the contributed happiness value of the child.
4. Eventually return the accumulated result as the output of maximizing the sum of happiness values.
Time Complexity Analysis:
-
The time complexity of sorting is O(n*log(n)) and n is the number of children.
-
When selecting k children, it is necessary to traverse up to k elements with a time complexity of O(k).
-
Thus, the total time complexity is O(n*log(n) + k).
Space complexity analysis:
- A constant level of extra space is required to perform the computation, so the total extra space complexity can be viewed as O(1).
The full Go code is below:
package main
import (
"fmt"
"slices"
)
func maximumHappinessSum(happiness []int, k int) (ans int64) {
(happiness, func(a, b int) int { return b - a })
for i, x := range happiness[:k] {
if x <= i {
break
}
ans += int64(x - i)
}
return
}
func main() {
happiness := []int{1, 2, 3}
k := 2
(maximumHappinessSum(happiness, k))
}
The full Rust code is below:
fn maximum_happiness_sum(happiness: &mut Vec<i32>, k: usize) -> i64 {
happiness.sort_by(|a, b| (a));
let mut ans: i64 = 0;
for (i, &x) in ().take(k).enumerate() {
if x <= i as i32 {
break;
}
ans += x as i64 - i as i64;
}
ans
}
fn main() {
let mut happiness = vec![1, 2, 3];
let k = 2;
println!("{}", maximum_happiness_sum(&mut happiness, k));
}
