#3075

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Maximize Happiness of Selected Children

Maximize the happiness of selected children by choosing the k happiest ones and applying greedy strategies to minimize losses.

array greedy sorting

Problem Statement

You are given an array happiness representing the happiness values of n children, and a positive integer k. Your task is to select k children in a way that maximizes their total happiness. When selecting a child, the happiness of all other children that have not been selected decreases by 1 (but never goes below 0).

The challenge lies in choosing the children optimally, considering that each non-selected child’s happiness decreases at the same rate. A greedy strategy must be employed to select the k largest happiness values, factoring in the incremental decrements for each pick.

Examples

Example 1

Input: happiness = [1,2,3], k = 2

Output: 4

We can pick 2 children in the following way:

Pick the child with the happiness value == 3. The happiness value of the remaining children becomes [0,1].
Pick the child with the happiness value == 1. The happiness value of the remaining child becomes [0]. Note that the happiness value cannot become less than 0. The sum of the happiness values of the selected children is 3 + 1 = 4.

Example 2

Input: happiness = [1,1,1,1], k = 2

Output: 1

We can pick 2 children in the following way:

Pick any child with the happiness value == 1. The happiness value of the remaining children becomes [0,0,0].
Pick the child with the happiness value == 0. The happiness value of the remaining child becomes [0,0]. The sum of the happiness values of the selected children is 1 + 0 = 1.

Example 3

Input: happiness = [2,3,4,5], k = 1

Output: 5

We can pick 1 child in the following way:

Pick the child with the happiness value == 5. The happiness value of the remaining children becomes [1,2,3]. The sum of the happiness values of the selected children is 5.

Constraints

1 <= n == happiness.length <= 2 * 105
1 <= happiness[i] <= 108
1 <= k <= n

Solution Approach

Greedy Selection

The most straightforward approach is to greedily select the children with the highest happiness values. After picking each child, decrease the happiness of the remaining ones, but ensure that the decrement never pushes a value below zero.

Sorting for Maximum Values

Sort the array in descending order. After sorting, you can easily pick the first k children. This ensures that you’re picking the largest values, minimizing the impact of the decrements on the happiness of the remaining children.

Validating Invariant

The key invariant is that the remaining children’s happiness decreases at the same rate. Therefore, by picking the largest values first, you are reducing the impact of future decrements on the total happiness score.

Complexity Analysis

Metric	Value
Time	O(n \cdot \log n + k \cdot \log n)
Space	O(n)

The time complexity is dominated by the sorting step, which takes O(n * log n), where n is the number of children. The decrement operation takes linear time, making the overall complexity O(n * log n + k * log n). Space complexity is O(n) for storing the array of happiness values.

What Interviewers Usually Probe

The candidate selects the top k happiness values correctly.
The candidate shows clear reasoning behind the greedy selection approach.
The candidate effectively manages the invariant regarding the decrementing happiness values.

Common Pitfalls or Variants

Common pitfalls

Failing to sort the array before making selections can lead to suboptimal choices.
Not ensuring that happiness values don’t go below zero after decrements.
Misunderstanding the problem's greedy approach by picking the wrong children.

Follow-up variants

What if there are multiple children with the same happiness value?
What if we select fewer than k children?
How would the problem change if there was no happiness decrement after each selection?

FAQ

What is the greedy approach for maximizing happiness in the Maximize Happiness of Selected Children problem?

The greedy approach involves sorting the happiness values in descending order and selecting the top k values. This minimizes the impact of happiness decrements.

Why is sorting important in the Maximize Happiness of Selected Children problem?

Sorting allows you to easily pick the highest happiness values first, which ensures you are selecting the most optimal children for maximizing happiness.

How does the decrementing of happiness values affect the solution?

The decrementing of happiness values affects the remaining children equally, making it crucial to select children with the highest initial happiness to minimize losses.

Can a dynamic programming approach be applied to this problem?

This problem is best solved using a greedy strategy, as it’s a simpler and more efficient way to handle the selection process with decreasing happiness values.

What’s the time complexity of the Maximize Happiness of Selected Children problem?

The time complexity is O(n * log n + k * log n), dominated by the sorting step, where n is the number of children.

terminal

Solution

Solution 1: Greedy + Sorting

To maximize the sum of happiness values, we should prioritize selecting children with higher happiness values. Therefore, we can sort the children in descending order by happiness value, and then select $k$ children in sequence. For the current $i$-th child, the happiness value obtained is $\max(\textit{happiness}[i] - i, 0)$. Finally, return the sum of happiness values of these $k$ children.

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class Solution:
    def maximumHappinessSum(self, happiness: List[int], k: int) -> int:
        happiness.sort(reverse=True)
        ans = 0
        for i, x in enumerate(happiness[:k]):
            x -= i
            ans += max(x, 0)
        return ans

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