BEATS 100% SIMPLE DART solution with EXPLANATION
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires us to select k children from a queue such that the sum of their ha
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires us to select k children from a queue such that the sum of their happiness values is maximized. Each time a child is selected, the happiness value of all the unselected children decreases by 1. To maximize the sum of happiness values, we should always select the child with the highest current happiness value.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach is to first sort the happiness values in descending order. Then, iterate over the sorted happiness values. For each happiness value, check if it minus the number of already taken children is greater than 0. If it is, add it to the sum and increment the number of taken children. If it\'s not, return the sum. Continue this process until all k children have been taken or there are no more children left.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity is O(n log n), where n is the number of children. This is because we are sorting the happiness values, which takes O(n log n) time.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is O(1), as we are not using any additional space that scales with the input size. The variables `taken`, `ans`, and `i` use constant space.\n# Code\n```\nclass Solution {\n  int maximumHappinessSum(List<int> happiness, int k) {\n    // Initialize the number of taken children and the sum of happiness values\n    int taken = 0;\n    int ans=0;\n    int i=0;\n\n    // Sort the happiness values in descending order\n    happiness.sort((a,b)=>b.compareTo(a));\n\n    // Iterate over the sorted happiness values\n    while(taken<k && i<happiness.length){\n        // If the current happiness value minus the number of taken children is less than or equal to 0, return the sum\n        if(happiness[i]-taken <= 0) return ans;\n\n        // Add the current happiness value minus the number of taken children to the sum\n        ans+=happiness[i]-taken;\n\n        // Increment the number of taken children\n        taken++;\n\n        // Move to the next happiness value\n        i++;\n    }\n\n    // Return the sum of happiness values\n    return ans;\n  }\n}\n```
['Array', 'Greedy', 'Sorting', 'Dart']