Table of Contents
Source: https://leetcode.com/problems/maximize-happiness-of-selected-children/description/
Description: Maximize Happiness of Selected Children
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
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<strong>Input:</strong> happiness = [1,2,3], k = 2 <strong>Output:</strong> 4 <strong>Explanation:</strong> 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
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<strong>Input:</strong> happiness = [1,1,1,1], k = 2 <strong>Output:</strong> 1 <strong>Explanation:</strong> 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
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<strong>Input:</strong> happiness = [2,3,4,5], k = 1 <strong>Output:</strong> 5 <strong>Explanation:</strong> 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 <= n
Solution
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class Solution { public long maximumHappinessSum(int[] happiness, int k) { //1. sort the array Arrays.sort(happiness); int n = happiness.length; long sum = 0; //to keep trck of the turns int itr = 0; //start from the last child with the happiness greater than all the other childs for(int i=n-1;i>=0;i--){ //if k is 0 return sum if(k==0){ return sum; } if(happiness[i]-itr > 0){ sum+= happiness[i] - itr; //increment turn itr++; } else { return sum; } k--; } return sum; } } |
Time Complexity
O(nlogn), where n is the size of the array happiness
Space Complexity
O(log(n)), where n is the size of the array happiness
Note: in java sort is implemented as a variant of quicksort which has space complexity of O(Log (n))
In the case of objects, java is going to use a hybrid between mergesort and insertion sort , called TimSort, which has O(n) space complexity. For primitive types, java will use a dual pivot variation of quicksort, which has O(log(n)) space complexity.