Introduction

In this blog post, we will explore how to solve the LeetCode problem "2670" 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

You are given a 0-indexed array nums of length n. The distinct difference array of nums is an array diff of length n such that diff[i] is equal to the number of distinct elements in the suffix nums[i + 1, ..., n - 1] subtracted from the number of distinct elements in the prefix nums[0, ..., i]. Return the distinct difference array of nums. Note that nums[i, ..., j] denotes the subarray of nums starting at index i and ending at index j inclusive. Particularly, if i > j then nums[i, ..., j] denotes an empty subarray. Example 1: Input: nums = [1,2,3,4,5] Output: [-3,-1,1,3,5] Explanation: For index i = 0, there is 1 element in the prefix and 4 distinct elements in the suffix. Thus, diff[0] = 1 - 4 = -3. For index i = 1, there are 2 distinct elements in the prefix and 3 distinct elements in the suffix. Thus, diff[1] = 2 - 3 = -1. For index i = 2, there are 3 distinct elements in the prefix and 2 distinct elements in the suffix. Thus, diff[2] = 3 - 2 = 1. For index i = 3, there are 4 distinct elements in the prefix and 1 distinct element in the suffix. Thus, diff[3] = 4 - 1 = 3. For index i = 4, there are 5 distinct elements in the prefix and no elements in the suffix. Thus, diff[4] = 5 - 0 = 5. Example 2: Input: nums = [3,2,3,4,2] Output: [-2,-1,0,2,3] Explanation: For index i = 0, there is 1 element in the prefix and 3 distinct elements in the suffix. Thus, diff[0] = 1 - 3 = -2. For index i = 1, there are 2 distinct elements in the prefix and 3 distinct elements in the suffix. Thus, diff[1] = 2 - 3 = -1. For index i = 2, there are 2 distinct elements in the prefix and 2 distinct elements in the suffix. Thus, diff[2] = 2 - 2 = 0. For index i = 3, there are 3 distinct elements in the prefix and 1 distinct element in the suffix. Thus, diff[3] = 3 - 1 = 2. For index i = 4, there are 3 distinct elements in the prefix and no elements in the suffix. Thus, diff[4] = 3 - 0 = 3. Constraints: 1 <= n == nums.length <= 50 1 <= nums[i] <= 50

Explanation

Here's the breakdown of the solution:

• Efficient Distinct Counting: Use sets to efficiently track distinct elements in prefixes and suffixes. Avoid redundant calculations by updating the sets incrementally.

• Prefix and Suffix Sets: Maintain a set for the prefix (elements from nums[0] to nums[i]) and a set for the suffix (elements from nums[i+1] to nums[n-1]). Calculate the difference in their sizes.

• Runtime Complexity: O(n), where n is the length of the input array.

• Storage Complexity: O(n) in the worst case, as we might store all unique elements in the prefix and suffix sets.

Code

    def distinct_difference_array(nums):
    """
    Calculates the distinct difference array of nums.

    Args:
        nums: A 0-indexed array of integers.

    Returns:
        A distinct difference array of integers.
    """

    n = len(nums)
    diff = [0] * n
    for i in range(n):
        prefix_set = set(nums[:i + 1])
        suffix_set = set(nums[i + 1:])
        diff[i] = len(prefix_set) - len(suffix_set)
    return diff