3075. Maximize Happiness of Selected Children
Problem Description
In this problem, we are presented with n
children standing in a queue, each with a certain "happiness" value. We aim to maximize the total happiness by selecting k
children. Notably, each time a child is chosen, the happiness value of every unselected child decreases by 1
, if their happiness is positive. The challenge is that as we select children, the unselected ones become less happy, so we need to carefully pick the sequence to make the most out of the happiness we can gather. The goal is to devise a strategy to get the maximum sum of happiness values by selecting k
children from the queue.
Intuition
To maximize the total happiness from selecting k
children, we need to grab the opportunity of the highest happiness values first. One optimal strategy is to prioritize children with higher happiness values before their potential happiness diminishes due to the selection of others. This calls for a greedy approach where sorting comes to play a vital role.
By sorting the children in descending order based on their happiness, we ensure that we select the happiest child first. Remember, each time we pick a child, subsequent children's happiness reduces by 1
ābut only if they're still positive. So, we greedily select from the top, decrementing the happiness value by the number of children already selected, symbolized by i
.
The solution approach involves:
- Sorting the children based on their happiness values in descending order.
- Iteratively selecting
k
children from the sorted list. - For each selected child, computing the effective happiness as the current happiness minus the count of previously selected children.
- Ensuring that if decrementing makes the happiness value negative, we consider it as zero since happiness cannot be negative by the problem's constraint.
- Summing up all the effective happiness values of the selected children to achieve the maximum sum.
Implementing this approach yields the maximum sum of happiness we can obtain from the given queue of children.
Solution Approach
To implement the solution effectively, we use the following algorithmic steps and Python built-in features aligned with the Greedy approach and the specified behavior of the problem:
-
Sorting: We start by sorting the
happiness
array in descending order. This is crucial because we want to select children with higher happiness values first, as each selection will decrease the potential happiness we can collect from the remaining children.happiness.sort(reverse=True)
By sorting in reverse order, we guarantee that
happiness[0]
holds the highest value,happiness[1]
the second-highest, and so on. -
Iterative Selection: We iterate over the first
k
children in the sortedhappiness
array. This is because, according to our Greedy strategy, these will be the children from whom we can extract the most happiness before their values start to decrease.for i, x in enumerate(happiness[:k]):
Here,
enumerate
is a built-in Python function that provides both the indexi
and the valuex
during the iteration. The indexi
represents how many children have been previously selected and thus by how much the currently considered child's happiness will be reduced. -
Happiness Calculation and Summation:
-
We calculate the adjusted happiness for each selected child. The adjustment is based on the assumption that each child's happiness decreases by
1
for every child already selected, which corresponds to the indexi
. -
We use the
max
function to ensure that we do not consider negative happiness values because the problem states that happiness cannot be negative even after decrementing.
x -= i ans += max(x, 0)
The adjusted happiness (
x - i
) is added to theans
, which accumulates the total happiness over the selection ofk
children. -
-
Returning the Result: After processing the
k
children, we have the maximum sum of happiness inans
, which we return as the result.return ans
Altogether, the algorithm uses sorting and a single pass through the first k
elements of the sorted array to compute the maximum sum of happiness. The Greedy approach ensures that, at each step, we make the locally optimal choice by picking the child with the maximum possible happiness at that turn.
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Start EvaluatorExample Walkthrough
Let's walk through an example to demonstrate the solution approach. Suppose we have n = 5
children with happiness values given by the array happiness = [5, 3, 1, 2, 4]
, and we want to maximize the total happiness by selecting k = 3
children from the queue.
Here are the steps weāll follow, as per the solution approach:
Step 1: Sorting
First, we sort the happiness
array in descending order:
happiness.sort(reverse=True) # [5, 4, 3, 2, 1]
Now, the happiness
array becomes [5, 4, 3, 2, 1]
.
Step 2: Iterative Selection
Next, we iterate over the first k
children in the sorted happiness
array. Applying the Greedy strategy, weāll pick the children with the highest happiness first:
for i, x in enumerate(happiness[:3]): # i goes from 0 to 2, x will be 5, then 4, then 3
Step 3: Happiness Calculation and Summation
On each iteration, we calculate the effective happiness (x - i
) and ensure it does not go below zero:
- First iteration (
i = 0
,x = 5
):x - i
is5 - 0
, which equals5
. So we add5
to the total happiness. - Second iteration (
i = 1
,x = 4
):x - i
is4 - 1
, which equals3
. We add3
to the total happiness. - Third iteration (
i = 2
,x = 3
):x - i
is3 - 2
, which equals1
. We add1
to the total happiness.
The ans
keeps accumulating these values, and we use the max
function to add only non-negative values:
ans = 0
ans += max(5 - 0, 0) # ans = 5
ans += max(4 - 1, 0) # ans = 5 + 3 = 8
ans += max(3 - 2, 0) # ans = 8 + 1 = 9
Step 4: Returning the Result
After processing k = 3
children, we have the total happiness:
return ans # returns 9
The resulting maximum sum of happiness values for selecting k = 3
children is 9
. The selection process prioritized children with the highest happiness value first and adjusted for each subsequent selection. This example confirms that our Greedy approach effectively maximizes the total happiness achieved.
Solution Implementation
1class Solution:
2 def maximumHappinessSum(self, happiness: List[int], k: int) -> int:
3 # Sort the happiness values in descending order.
4 happiness.sort(reverse=True)
5 max_happiness = 0 # Initialize the maximum happiness sum.
6
7 # Loop through the first k elements after sorting.
8 for i in range(k):
9 # Since a personās happiness value can decrease with each additional person,
10 # we subtract the current index from the happiness value,
11 # however, we should not go below 0.
12 decreased_happiness = happiness[i] - i
13 if decreased_happiness > 0:
14 max_happiness += decreased_happiness
15
16 # Return the sum of the maximized happiness values.
17 return max_happiness
18
1class Solution {
2 public long maximumHappinessSum(int[] happiness, int k) {
3 // Sort the array so that we can easily pick the largest elements
4 Arrays.sort(happiness);
5
6 // Initialize the sum of happiness to 0
7 long totalHappiness = 0;
8
9 // Iterate through the array starting from the end to get the largest values
10 for (int i = 0; i < k; ++i) {
11 // Calculate the current happiness after decrementing based on index i
12 // Using n - i - 1 to pick the k largest elements in a sorted array
13 int currentHappiness = happiness[happiness.length - i - 1] - i;
14
15 // Sum only the positive values of current happiness after the decrement
16 totalHappiness += Math.max(currentHappiness, 0);
17 }
18
19 // Return the total happiness calculation
20 return totalHappiness;
21 }
22}
23
1#include <vector>
2#include <algorithm> // Include algorithm header for std::sort
3
4class Solution {
5public:
6 // Function to calculate the maximum happiness sum using the provided happiness vector and integer k
7 long long maximumHappinessSum(vector<int>& happiness, int k) {
8 // Sort the happiness vector in non-increasing order
9 sort(happiness.rbegin(), happiness.rend());
10
11 // Initialize a variable to store the accumulated maximum happiness sum
12 long long maxHappinessSum = 0;
13
14 // Iterate over the first k elements of the sorted vector
15 for (int i = 0, n = happiness.size(); i < k; ++i) {
16 // Calculate the modified happiness value by subtracting the index
17 int modifiedHappiness = happiness[i] - i;
18
19 // If the modified happiness value is positive, add it to the sum
20 // Otherwise, add zero (do not decrease the sum)
21 maxHappinessSum += std::max(modifiedHappiness, 0);
22 }
23
24 // Return the accumulated maximum happiness sum
25 return maxHappinessSum;
26 }
27};
28
1// This function calculates the maximum happiness sum from an array of happiness points.
2// It considers the 'k' people with the highest happiness points after applying a specific penalty.
3// The penalty subtracts the person's index from their happiness points (starting from 0).
4
5function maximumHappinessSum(happiness: number[], k: number): number {
6 // Sort the happiness array in descending order.
7 happiness.sort((a, b) => b - a);
8 // Initialize a variable to keep track of the total happiness sum.
9 let totalHappiness = 0;
10
11 // Loop through the first 'k' elements in the sorted happiness array.
12 for (let i = 0; i < k; ++i) {
13 // Calculate the penalty by subtracting the index (0-based) from the happiness points.
14 const penalizedHappiness = happiness[i] - i;
15 // Add the greater of penalizedHappiness or 0 to the totalHappiness sum
16 // to ensure that negative values do not reduce the overall happiness.
17 totalHappiness += Math.max(penalizedHappiness, 0);
18 }
19 // Return the total happiness sum after processing the first 'k' elements.
20 return totalHappiness;
21}
22
Time and Space Complexity
The time complexity of the maximumHappinessSum
function is composed of two parts: sorting the happiness
list and iterating over a slice of this list.
- Sorting the list requires
O(n * log n)
time, wheren
is the length of thehappiness
list. The Pythonsort()
method uses the Timsort algorithm, which has this complexity. - Iterating over the first
k
elements of the sorted list takesO(k)
time, because we're only looking at a subset of the list, which containsk
elements.
Adding these two components together, the total time complexity is O(n * log n + k)
.
As for the space complexity, the sort operation can be done in-place, but Timsort requires additional temporary space for its operation. This temporary space is O(log n)
because of the way the algorithm divides the list and merges sorted sublists. There is no additional significant space usage since only a few extra variables are created, and these do not depend on the size of the input list.
Learn more about how to find time and space complexity quickly using problem constraints.
Which of these properties could exist for a graph but not a tree?
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