Introduction

In this blog post, we will explore how to solve the LeetCode problem "2799" 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 an array nums consisting of positive integers. We call a subarray of an array complete if the following condition is satisfied: The number of distinct elements in the subarray is equal to the number of distinct elements in the whole array. Return the number of complete subarrays. A subarray is a contiguous non-empty part of an array. Example 1: Input: nums = [1,3,1,2,2] Output: 4 Explanation: The complete subarrays are the following: [1,3,1,2], [1,3,1,2,2], [3,1,2] and [3,1,2,2]. Example 2: Input: nums = [5,5,5,5] Output: 10 Explanation: The array consists only of the integer 5, so any subarray is complete. The number of subarrays that we can choose is 10. Constraints: 1 <= nums.length <= 1000 1 <= nums[i] <= 2000

Explanation

Here's a breakdown of the solution strategy and the Python code:

• Core Idea: Iterate through the array using a sliding window approach. For each starting position of the window, expand the window to the right until it becomes a "complete" subarray. Once complete, count all the subarrays starting from that position.
• Distinct Element Tracking: Use a dictionary to efficiently track the frequency of elements within the current window.

• Optimization: Shrink the window from the left while it remains a complete subarray, maximizing the count of complete subarrays starting from the initial window start position.

• Time & Space Complexity: O(n), O(n) where n is the length of the input array, nums.

Code

    def countCompleteSubarrays(nums):
    n = len(nums)
    distinct_count = len(set(nums))
    count = 0

    for i in range(n):
        freq = {}
        distinct_in_window = 0
        for j in range(i, n):
            if nums[j] not in freq:
                freq[nums[j]] = 0
                distinct_in_window += 1
            freq[nums[j]] += 1

            if distinct_in_window == distinct_count:
                count += (n - j)
                break  # Optimization: Once complete, count and move to next i

    return count