Solution
Intuition\nThe problem is asking to maximize the sum of happiness by choosing k elements from the given array. The happiness of an athlete decreases by 1 for ea
# Intuition\nThe problem is asking to maximize the sum of happiness by choosing `k` elements from the given array. The happiness of an athlete decreases by 1 for each athlete that has a higher score. So, we need to choose the `k` largest elements from the array to maximize the sum.\n\n# Approach\n- We first sort the array in ascending order.\n- Then, we iterate from the end of the array (which contains the largest elements) towards the beginning.\n- For each element, we check if subtracting the current rank (`j`) from the happiness score will result in a non-positive number or if we have already chosen `k` elements. If either of these conditions is true, we break the loop.\n- Otherwise, we add the adjusted happiness score (happiness score minus rank) to the total sum and increment the rank.\n\n# Complexity\n- Time complexity: The time complexity is $$O(n \\log n)$$ due to the sorting operation, where `n` is the length of the input array.\n- Space complexity: The space complexity is $$O(1)$$ as we are not using any additional space that scales with the input size.\n\n# Code\nThe provided code correctly implements the described approach. Here it is again for reference:\n\n```java\nclass Solution {\n    public long maximumHappinessSum(int[] happiness, int k) {\n        long ans = 0;\n        int j = 0;\n        int n = happiness.length;\n        Arrays.sort(happiness);\n        for(int i=n-1;i>=0;i--){\n            if(happiness[i]-j <= 0 || j==k){\n                break;\n            }\n            ans = ans + (happiness[i]-j);\n            j++;\n        }\n\n        return ans;\n    }\n}\n```
['Java']