LeetCode 3478: Choose K Elements With Maximum Sum Solution

Master LeetCode problem 3478 (Choose K Elements With Maximum Sum), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3478. Choose K Elements With Maximum Sum

Medium

Array Sorting Heap (Priority Queue)

Problem Explanation

Explanation:

To solve this problem, we can iterate over the elements of nums1 and for each element, find the indices in nums1 where the corresponding value is less than the current element. Then, we can choose up to k maximum values from nums2 at these indices to maximize the total sum. We can achieve this by sorting the indices based on nums2 values and selecting the top k values. Finally, we calculate the sum for each element in nums1.

Algorithm:

Create a mapping from nums1 values to indices for quick lookup.
Iterate over nums1 elements.
For each element, find the indices in nums1 where the value is less than the current element.
Sort these indices based on nums2 values in descending order.
Choose up to k maximum values from nums2 at these indices.
Calculate the sum for the current element and store it in the answer array.
Return the answer array.

Time Complexity:

The time complexity of this approach is O(n log n) due to sorting for each element in nums1.

Space Complexity:

The space complexity is O(n) for storing the indices mapping and the answer array.

Solution Code

import java.util.*;

class Solution {
    public int[] maxSumOfKElements(int[] nums1, int[] nums2, int k) {
        int n = nums1.length;
        Map<Integer, List<Integer>> indexMap = new HashMap<>();
        for (int i = 0; i < n; i++) {
            indexMap.computeIfAbsent(nums1[i], key -> new ArrayList<>()).add(i);
        }

        int[] answer = new int[n];
        for (int i = 0; i < n; i++) {
            List<Integer> indices = indexMap.getOrDefault(nums1[i], new ArrayList<>());
            List<Integer> selectedIndices = new ArrayList<>();
            for (int index : indices) {
                if (nums1[index] < nums1[i]) {
                    selectedIndices.add(index);
                }
            }
            Collections.sort(selectedIndices, (a, b) -> Integer.compare(nums2[b], nums2[a]));

            int sum = 0;
            for (int j = 0; j < Math.min(k, selectedIndices.size()); j++) {
                sum += nums2[selectedIndices.get(j)];
            }
            answer[i] = sum;
        }

        return answer;
    }
}

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