Introduction

In this blog post, we will explore how to solve the LeetCode problem "3134" 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 integer array nums. The uniqueness array of nums is the sorted array that contains the number of distinct elements of all the subarrays of nums. In other words, it is a sorted array consisting of distinct(nums[i..j]), for all 0 <= i <= j < nums.length. Here, distinct(nums[i..j]) denotes the number of distinct elements in the subarray that starts at index i and ends at index j. Return the median of the uniqueness array of nums. Note that the median of an array is defined as the middle element of the array when it is sorted in non-decreasing order. If there are two choices for a median, the smaller of the two values is taken. Example 1: Input: nums = [1,2,3] Output: 1 Explanation: The uniqueness array of nums is [distinct(nums[0..0]), distinct(nums[1..1]), distinct(nums[2..2]), distinct(nums[0..1]), distinct(nums[1..2]), distinct(nums[0..2])] which is equal to [1, 1, 1, 2, 2, 3]. The uniqueness array has a median of 1. Therefore, the answer is 1. Example 2: Input: nums = [3,4,3,4,5] Output: 2 Explanation: The uniqueness array of nums is [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3]. The uniqueness array has a median of 2. Therefore, the answer is 2. Example 3: Input: nums = [4,3,5,4] Output: 2 Explanation: The uniqueness array of nums is [1, 1, 1, 1, 2, 2, 2, 3, 3, 3]. The uniqueness array has a median of 2. Therefore, the answer is 2. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 105

Explanation

• Binary Search for Median: Instead of generating the entire uniqueness array and sorting it (which would be inefficient), use binary search to find the median value directly. The core idea is to guess a potential median value and then count how many subarrays have a distinct count less than or equal to that guessed median.
  • Counting Subarrays with Distinct Count <= Guess: For each guess in the binary search, efficiently count the number of subarrays whose distinct element count is less than or equal to the guess. This counting involves iterating through all possible subarrays, using a sliding window and hash map to track distinct element counts.
  • Binary Search Convergence: Adjust the binary search range (low and high) based on the count obtained in the previous step. If the count is less than half the size of the uniqueness array, the guess is too low; otherwise, the guess is too high. This converges to the median value.

• Runtime Complexity: O(N² log N), where N is the length of nums. The O(N²) comes from the sliding window and checking all subarrays, and O(log N) comes from the binary search. Storage Complexity: O(N), due to the hash map used within the sliding window.

Code

    def solve():
    def count_subarrays(nums, target):
        n = len(nums)
        count = 0
        for i in range(n):
            distinct_count = 0
            freq = {}
            for j in range(i, n):
                if nums[j] not in freq:
                    freq[nums[j]] = 0
                if freq[nums[j]] == 0:
                    distinct_count += 1
                freq[nums[j]] += 1
                if distinct_count <= target:
                    count += 1
        return count

    nums = [int(x) for x in input().split(",")]
    n = len(nums)
    total_subarrays = n * (n + 1) // 2
    low = 1
    high = n
    ans = 0

    while low <= high:
        mid = (low + high) // 2
        if count_subarrays(nums, mid) >= (total_subarrays + 1) // 2:
            ans = mid
            high = mid - 1
        else:
            low = mid + 1

    print(ans)

solve()