Introduction

In this blog post, we will explore how to solve the LeetCode problem "1589" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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

We have an array of integers, nums, and an array of requests where requests[i] = [starti, endi]. The ith request asks for the sum of nums[starti] + nums[starti + 1] + ... + nums[endi - 1] + nums[endi]. Both starti and endi are 0-indexed. Return the maximum total sum of all requests among all permutations of nums. Since the answer may be too large, return it modulo 109 + 7. Example 1: Input: nums = [1,2,3,4,5], requests = [[1,3],[0,1]] Output: 19 Explanation: One permutation of nums is [2,1,3,4,5] with the following result: requests[0] -> nums[1] + nums[2] + nums[3] = 1 + 3 + 4 = 8 requests[1] -> nums[0] + nums[1] = 2 + 1 = 3 Total sum: 8 + 3 = 11. A permutation with a higher total sum is [3,5,4,2,1] with the following result: requests[0] -> nums[1] + nums[2] + nums[3] = 5 + 4 + 2 = 11 requests[1] -> nums[0] + nums[1] = 3 + 5 = 8 Total sum: 11 + 8 = 19, which is the best that you can do. Example 2: Input: nums = [1,2,3,4,5,6], requests = [[0,1]] Output: 11 Explanation: A permutation with the max total sum is [6,5,4,3,2,1] with request sums [11]. Example 3: Input: nums = [1,2,3,4,5,10], requests = [[0,2],[1,3],[1,1]] Output: 47 Explanation: A permutation with the max total sum is [4,10,5,3,2,1] with request sums [19,18,10]. Constraints: n == nums.length 1 <= n <= 105 0 <= nums[i] <= 105 1 <= requests.length <= 105 requests[i].length == 2 0 <= starti <= endi < n

Explanation

Here's an efficient solution to the problem:

• Frequency Counting: Calculate how many times each index in nums is included in the requests. This will help determine which numbers should be placed at the most frequently requested indices.
• Sorting: Sort the nums array and the frequency array. The largest numbers should be placed at the most frequently requested indices to maximize the total sum.

• Calculate Total Sum: Calculate the total sum of requests using the optimized permutation.

• Runtime Complexity: O(n log n + m log m), where n is the length of nums and m is the length of requests. This is due to the sorting operations. Storage Complexity: O(n) due to the frequency array.

Code

    def maxSumRangeQuery(nums, requests):
    n = len(nums)
    freq = [0] * n
    for start, end in requests:
        freq[start] += 1
        if end + 1 < n:
            freq[end + 1] -= 1

    for i in range(1, n):
        freq[i] += freq[i - 1]

    nums.sort()
    freq.sort()

    total_sum = 0
    for i in range(n):
        total_sum = (total_sum + nums[i] * freq[i]) % (10**9 + 7)

    return total_sum