# Solving Leetcode Interviews in Seconds with AI: Maximum Sum Queries


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2736" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given two 0-indexed integer arrays nums1 and nums2, each of length n, and a 1-indexed 2D array queries where queries[i] = [xi, yi]. For the ith query, find the maximum value of nums1[j] + nums2[j] among all indices j (0 <= j < n), where nums1[j] >= xi and nums2[j] >= yi, or -1 if there is no j satisfying the constraints. Return an array answer where answer[i] is the answer to the ith query.   Example 1:  Input: nums1 = [4,3,1,2], nums2 = [2,4,9,5], queries = [[4,1],[1,3],[2,5]] Output: [6,10,7] Explanation:  For the 1st query xi = 4 and yi = 1, we can select index j = 0 since nums1[j] >= 4 and nums2[j] >= 1. The sum nums1[j] + nums2[j] is 6, and we can show that 6 is the maximum we can obtain.  For the 2nd query xi = 1 and yi = 3, we can select index j = 2 since nums1[j] >= 1 and nums2[j] >= 3. The sum nums1[j] + nums2[j] is 10, and we can show that 10 is the maximum we can obtain.   For the 3rd query xi = 2 and yi = 5, we can select index j = 3 since nums1[j] >= 2 and nums2[j] >= 5. The sum nums1[j] + nums2[j] is 7, and we can show that 7 is the maximum we can obtain.  Therefore, we return [6,10,7].  Example 2:  Input: nums1 = [3,2,5], nums2 = [2,3,4], queries = [[4,4],[3,2],[1,1]] Output: [9,9,9] Explanation: For this example, we can use index j = 2 for all the queries since it satisfies the constraints for each query.  Example 3:  Input: nums1 = [2,1], nums2 = [2,3], queries = [[3,3]] Output: [-1] Explanation: There is one query in this example with xi = 3 and yi = 3. For every index, j, either nums1[j] < xi or nums2[j] < yi. Hence, there is no solution.     Constraints:  nums1.length == nums2.length  n == nums1.length  1 <= n <= 105 1 <= nums1[i], nums2[i] <= 109  1 <= queries.length <= 105 queries[i].length == 2 xi == queries[i][1] yi == queries[i][2] 1 <= xi, yi <= 109  

	# Explanation
	Here's an efficient solution to the problem, along with explanations:

**High-Level Approach:**

*   **Sort and Preprocess:** Sort the indices based on `nums1` in descending order. This allows us to efficiently iterate through potential candidates that satisfy the `nums1[j] >= xi` condition.
*   **Maintain a Max Heap:** Use a max heap (priority queue) to store `nums1[j] + nums2[j]` for indices that satisfy `nums1[j] >= xi`. The heap is sorted by `nums2[j]` in descending order, ensuring that the element at the top always corresponds to the maximum `nums1[j] + nums2[j]` encountered so far that satisfies the `nums2[j] >= yi` condition.
*   **Process Queries Offline:** Sort the queries in descending order of `xi`. This ensures that the max heap only contains candidate indices that have `nums1[j] >= xi`.

**Complexity:**

*   **Runtime Complexity:** O(n log n + q log q + q log n), where n is the length of nums1 and nums2, and q is the length of queries.  Sorting takes O(n log n + q log q), and inserting/popping from the heap takes O(log n) time for each of the q queries.
*   **Storage Complexity:** O(n + q)

	
	# Code
	```python
	import heapq

def solve():
    nums1 = [4,3,1,2]
    nums2 = [2,4,9,5]
    queries = [[4,1],[1,3],[2,5]]
    print(max_sum_queries(nums1, nums2, queries))

def max_sum_queries(nums1, nums2, queries):
    n = len(nums1)
    indices = sorted(range(n), key=lambda i: nums1[i], reverse=True)
    queries_with_indices = sorted([(x, y, i) for i, (x, y) in enumerate(queries)], key=lambda q: q[0], reverse=True)
    
    answers = [-1] * len(queries)
    heap = []
    index_ptr = 0
    
    for xi, yi, query_index in queries_with_indices:
        # Add eligible indices to the heap
        while index_ptr < n and nums1[indices[index_ptr]] >= xi:
            j = indices[index_ptr]
            heapq.heappush(heap, (- (nums1[j] + nums2[j]), -nums2[j]))  # Negate for max heap
            index_ptr += 1
        
        # Find the maximum sum that satisfies nums2[j] >= yi
        while heap and -heap[0][1] < yi:
            heapq.heappop(heap)
        
        if heap:
            answers[query_index] = -heap[0][0]
            
    return answers
	```