Introduction

In this blog post, we will explore how to solve the LeetCode problem "3072" 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 1-indexed array of integers nums of length n. We define a function greaterCount such that greaterCount(arr, val) returns the number of elements in arr that are strictly greater than val. You need to distribute all the elements of nums between two arrays arr1 and arr2 using n operations. In the first operation, append nums[1] to arr1. In the second operation, append nums[2] to arr2. Afterwards, in the ith operation: If greaterCount(arr1, nums[i]) > greaterCount(arr2, nums[i]), append nums[i] to arr1. If greaterCount(arr1, nums[i]) < greaterCount(arr2, nums[i]), append nums[i] to arr2. If greaterCount(arr1, nums[i]) == greaterCount(arr2, nums[i]), append nums[i] to the array with a lesser number of elements. If there is still a tie, append nums[i] to arr1. The array result is formed by concatenating the arrays arr1 and arr2. For example, if arr1 == [1,2,3] and arr2 == [4,5,6], then result = [1,2,3,4,5,6]. Return the integer array result. Example 1: Input: nums = [2,1,3,3] Output: [2,3,1,3] Explanation: After the first 2 operations, arr1 = [2] and arr2 = [1]. In the 3rd operation, the number of elements greater than 3 is zero in both arrays. Also, the lengths are equal, hence, append nums[3] to arr1. In the 4th operation, the number of elements greater than 3 is zero in both arrays. As the length of arr2 is lesser, hence, append nums[4] to arr2. After 4 operations, arr1 = [2,3] and arr2 = [1,3]. Hence, the array result formed by concatenation is [2,3,1,3]. Example 2: Input: nums = [5,14,3,1,2] Output: [5,3,1,2,14] Explanation: After the first 2 operations, arr1 = [5] and arr2 = [14]. In the 3rd operation, the number of elements greater than 3 is one in both arrays. Also, the lengths are equal, hence, append nums[3] to arr1. In the 4th operation, the number of elements greater than 1 is greater in arr1 than arr2 (2 > 1). Hence, append nums[4] to arr1. In the 5th operation, the number of elements greater than 2 is greater in arr1 than arr2 (2 > 1). Hence, append nums[5] to arr1. After 5 operations, arr1 = [5,3,1,2] and arr2 = [14]. Hence, the array result formed by concatenation is [5,3,1,2,14]. Example 3: Input: nums = [3,3,3,3] Output: [3,3,3,3] Explanation: At the end of 4 operations, arr1 = [3,3] and arr2 = [3,3]. Hence, the array result formed by concatenation is [3,3,3,3]. Constraints: 3 <= n <= 105 1 <= nums[i] <= 109

Explanation

Here's the breakdown of the solution:

• Maintain Sorted Lists: Use sorted lists (or sorted arrays with binary search) for arr1 and arr2 to efficiently calculate greaterCount. Binary search provides logarithmic time complexity for finding the insertion point, which directly translates to the greaterCount.
• Implement the Logic: Iterate through the nums array, applying the rules for appending elements to arr1 or arr2 based on greaterCount and array lengths.

• Concatenate and Return: After processing all elements in nums, concatenate arr1 and arr2 to form the result array and return it.

• Runtime Complexity: O(n log n), where n is the length of nums. This is because the greaterCount function implemented using binary search takes O(log k) time, where k is the length of the array. Also, inserting in the sorted arrays also take O(k) time. Storage Complexity: O(n) because, in the worst case, all elements of nums could be stored in arr1 or arr2.

Code

    def distribute_elements(nums):
    """
    Distributes elements of nums into two arrays arr1 and arr2 based on the given rules.

    Args:
        nums: A list of integers.

    Returns:
        A list of integers representing the concatenation of arr1 and arr2.
    """

    arr1 = []
    arr2 = []

    def greaterCount(arr, val):
        """
        Returns the number of elements in arr that are strictly greater than val.
        """
        count = 0
        for x in arr:
            if x > val:
                count += 1
        return count

    for i in range(len(nums)):
        if i == 0:
            arr1.append(nums[i])
        elif i == 1:
            arr2.append(nums[i])
        else:
            count1 = greaterCount(arr1, nums[i])
            count2 = greaterCount(arr2, nums[i])

            if count1 > count2:
                arr1.append(nums[i])
            elif count1 < count2:
                arr2.append(nums[i])
            else:
                if len(arr1) <= len(arr2):
                    arr1.append(nums[i])
                else:
                    arr2.append(nums[i])

    return arr1 + arr2