LeetCode 3430: Maximum and Minimum Sums of at Most Size K Subarrays Solution

Master LeetCode problem 3430 (Maximum and Minimum Sums of at Most Size K Subarrays), a hard 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.

3430. Maximum and Minimum Sums of at Most Size K Subarrays

Hard

Array Math Stack Monotonic Stack

Problem Explanation

Explanation

To solve this problem, we can iterate through all subarrays with at most k elements and calculate the sum of the maximum and minimum elements in each subarray. We can keep track of the running total of these sums and return the final total.

Initialize two deques to store the indices of the elements in the current window with the maximum and minimum values.
Iterate through the array and for each element:
- Remove indices that are outside the current window from both deques.
- Add the current index to the end of the deque for maximum values.
- Add the current index to the end of the deque for minimum values.
- If the window size exceeds k, remove the front index from both deques.
- Calculate the sum of the maximum and minimum values in the current window and update the running total.
Return the final total.

Time complexity: O(n) Space complexity: O(k)

Solution Code

import java.util.*;

class Solution {
    public int maxMinSum(int[] nums, int k) {
        int total = 0;
        Deque<Integer> maxQueue = new LinkedList<>();
        Deque<Integer> minQueue = new LinkedList();

        for (int i = 0; i < nums.length; i++) {
            while (!maxQueue.isEmpty() && nums[i] >= nums[maxQueue.peekLast()]) {
                maxQueue.removeLast();
            }
            while (!minQueue.isEmpty() && nums[i] <= nums[minQueue.peekLast()]) {
                minQueue.removeLast();
            }

            maxQueue.addLast(i);
            minQueue.addLast(i);

            if (i - maxQueue.peekFirst() >= k) {
                maxQueue.removeFirst();
            }
            if (i - minQueue.peekFirst() >= k) {
                minQueue.removeFirst();
            }

            if (i >= k - 1) {
                total += nums[maxQueue.peekFirst()] + nums[minQueue.peekFirst()];
            }
        }

        return total;
    }
}

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