You are given an array happiness of length n, and a positive integer k.
There are n children standing in a queue, where the ith child has happiness value happiness[i]. You want to select k children from these n children in k turns.
In each turn, when you select a child, the happiness value of all the children that have not been selected till now decreases by 1. Note that the happiness value cannot become negative and gets decremented only if it is positive.
Return the maximum sum of the happiness values of the selected children you can achieve by selecting k children.
Example 1:
Input: happiness = [1,2,3], k = 2 Output: 4 Explanation: 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 Explanation: 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 Explanation: 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 * 1051 <= happiness[i] <= 1081 <= k <= nThis approach involves sorting the input data first and then finding the solution by traversing through the sorted data. This approach is generally straightforward and often leads to a solution by leveraging sorted order, which simplifies many problems, such as finding pairs or detecting duplicates.
This C program sorts the input array using the C standard library's qsort function and prints the sorted elements. You can replace the print operation with the logic required for your problem, like finding pairs or specific elements.
C++
Java
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Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) or O(n), depending on the usage of additional data structures.
This approach leverages a hash map to efficiently solve problems requiring quick lookups or to detect duplicates. This method is optimal for problems where you need to count occurrences or require O(1) average-time complexity for lookups.
This C example uses a simple hash map as an array to count occurrences of each number in the array. This method is useful for frequency counting or detecting duplicates.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n)
Space Complexity: O(n) for the hash map.
| Approach | Complexity |
|---|---|
| Sorting Approach | Time Complexity: O(n log n) due to sorting. |
| Hash Map Approach | Time Complexity: O(n) |
3075. Maximize Happiness of Selected Children | Greedy • Aryan Mittal • 1,687 views views
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