# Solving Leetcode Interviews in Seconds with AI: Sum of Imbalance Numbers of All Subarrays


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2763" 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
	> The imbalance number of a 0-indexed integer array arr of length n is defined as the number of indices in sarr = sorted(arr) such that:  0 <= i < n - 1, and sarr[i+1] - sarr[i] > 1  Here, sorted(arr) is the function that returns the sorted version of arr. Given a 0-indexed integer array nums, return the sum of imbalance numbers of all its subarrays. A subarray is a contiguous non-empty sequence of elements within an array.   Example 1:  Input: nums = [2,3,1,4] Output: 3 Explanation: There are 3 subarrays with non-zero imbalance numbers: - Subarray [3, 1] with an imbalance number of 1. - Subarray [3, 1, 4] with an imbalance number of 1. - Subarray [1, 4] with an imbalance number of 1. The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 3.   Example 2:  Input: nums = [1,3,3,3,5] Output: 8 Explanation: There are 7 subarrays with non-zero imbalance numbers: - Subarray [1, 3] with an imbalance number of 1. - Subarray [1, 3, 3] with an imbalance number of 1. - Subarray [1, 3, 3, 3] with an imbalance number of 1. - Subarray [1, 3, 3, 3, 5] with an imbalance number of 2.  - Subarray [3, 3, 3, 5] with an imbalance number of 1.  - Subarray [3, 3, 5] with an imbalance number of 1. - Subarray [3, 5] with an imbalance number of 1. The imbalance number of all other subarrays is 0. Hence, the sum of imbalance numbers of all the subarrays of nums is 8.    Constraints:  1 <= nums.length <= 1000 1 <= nums[i] <= nums.length  

	# Explanation
	Here's the breakdown of the solution:

*   **Iterate through all subarrays:** The core idea is to generate all possible subarrays of the input array `nums`.
*   **Calculate imbalance for each subarray:** For each subarray, sort it and then count the number of times the difference between consecutive elements is greater than 1. This gives the imbalance number for that subarray.
*   **Sum the imbalance numbers:** Accumulate the imbalance numbers of all subarrays to get the final result.

*   **Runtime Complexity:** O(n^2 log n), where n is the length of the input array `nums`. This is due to the nested loops for generating subarrays (O(n^2)) and sorting each subarray (O(n log n) in the worst case).
*   **Storage Complexity:** O(n) in the worst case, primarily due to the space used for storing the sorted subarray.

	
	# Code
	```python
	def sum_imbalance_numbers(nums):
    """
    Calculates the sum of imbalance numbers of all subarrays of a given array.

    Args:
        nums: A list of integers.

    Returns:
        The sum of imbalance numbers of all subarrays.
    """

    n = len(nums)
    total_imbalance = 0

    for i in range(n):
        for j in range(i, n):
            subarray = nums[i:j+1]
            sorted_subarray = sorted(subarray)
            imbalance = 0
            for k in range(len(sorted_subarray) - 1):
                if sorted_subarray[k+1] - sorted_subarray[k] > 1:
                    imbalance += 1
            total_imbalance += imbalance

    return total_imbalance
	```