# Solving Leetcode Interviews in Seconds with AI: Sum of Distances


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2615" 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 a 0-indexed integer array nums. There exists an array arr of length nums.length, where arr[i] is the sum of |i - j| over all j such that nums[j] == nums[i] and j != i. If there is no such j, set arr[i] to be 0. Return the array arr.   Example 1:  Input: nums = [1,3,1,1,2] Output: [5,0,3,4,0] Explanation:  When i = 0, nums[0] == nums[2] and nums[0] == nums[3]. Therefore, arr[0] = |0 - 2| + |0 - 3| = 5.  When i = 1, arr[1] = 0 because there is no other index with value 3. When i = 2, nums[2] == nums[0] and nums[2] == nums[3]. Therefore, arr[2] = |2 - 0| + |2 - 3| = 3.  When i = 3, nums[3] == nums[0] and nums[3] == nums[2]. Therefore, arr[3] = |3 - 0| + |3 - 2| = 4.  When i = 4, arr[4] = 0 because there is no other index with value 2.    Example 2:  Input: nums = [0,5,3] Output: [0,0,0] Explanation: Since each element in nums is distinct, arr[i] = 0 for all i.    Constraints:  1 <= nums.length <= 105 0 <= nums[i] <= 109    Note: This question is the same as  2121: Intervals Between Identical Elements. 

	# Explanation
	Here's the breakdown of the problem and the solution:

*   **High-Level Approach:**
    *   Use a dictionary to store the indices of each number in `nums`.
    *   Iterate through `nums`. For each element, retrieve the list of indices where the same number appears. Calculate the sum of absolute differences between the current index and all other indices in the list.
    *   Store the calculated sum in the result array `arr`.

*   **Complexity:**
    *   Runtime: O(n), where n is the length of `nums`.
    *   Storage: O(n), where n is the length of `nums`. This is because, in the worst case, all the elements in `nums` can be the same, and the dictionary might store all the indices.

	
	# Code
	```python
	def interval_between_identical_elements(nums):
    """
    Calculates the sum of absolute differences between indices of identical elements.

    Args:
        nums: A 0-indexed integer array.

    Returns:
        An array arr of the same length as nums, where arr[i] is the sum of |i - j|
        over all j such that nums[j] == nums[i] and j != i.
    """
    index_map = {}
    for i, num in enumerate(nums):
        if num not in index_map:
            index_map[num] = []
        index_map[num].append(i)

    arr = [0] * len(nums)
    for i in range(len(nums)):
        if nums[i] in index_map:
            indices = index_map[nums[i]]
            for j in indices:
                if i != j:
                    arr[i] += abs(i - j)

    return arr
	```