LeetCode 2736: Maximum Sum Queries Solution

Master LeetCode problem 2736 (Maximum Sum Queries), 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.

2736. Maximum Sum Queries

Hard

Array Binary Search Stack Binary Indexed Tree Segment Tree Sorting Monotonic Stack

Problem Explanation

Explanation:

To solve this problem efficiently, we can iterate through the queries and maintain two arrays max1 and max2 to store the maximum values of nums1[j] and nums2[j] for each index j. Then, we can iterate through the queries again and find the maximum sum for each query by checking the stored maximum values from max1 and max2.

Initialize max1 and max2 arrays of size n filled with -1.
Iterate through nums1 and nums2 to update max1 and max2.
Iterate through queries and check the conditions to calculate the maximum sum.

Time Complexity:

The time complexity of this approach is O(n) to preprocess the maximum values and O(q) to process the queries, where n is the length of the arrays and q is the number of queries. So, the overall time complexity is O(n + q).

Space Complexity:

The space complexity is O(n) for the two max1 and max2 arrays.

Solution Code

class Solution {
    public int[] maxSumQueries(int[] nums1, int[] nums2, int[][] queries) {
        int n = nums1.length;
        int[] max1 = new int[n];
        int[] max2 = new int[n];

        Arrays.fill(max1, -1);
        Arrays.fill(max2, -1);

        for (int i = 0; i < n; i++) {
            max1[i] = Math.max(max1[i], nums1[i]);
            max2[i] = Math.max(max2[i], nums2[i]);
        }

        int[] answer = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int xi = queries[i][0];
            int yi = queries[i][1];
            int maxSum = -1;

            for (int j = 0; j < n; j++) {
                if (nums1[j] >= xi && nums2[j] >= yi) {
                    maxSum = Math.max(maxSum, nums1[j] + nums2[j]);
                }
            }
            answer[i] = maxSum;
        }

        return answer;
    }
}

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