# Solving Leetcode Interviews in Seconds with AI: Maximum Sum Obtained of Any Permutation


	# 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](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
	> 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
	```python
	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
	```