Introduction

In this blog post, we will explore how to solve the LeetCode problem "954" 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

Given an integer array of even length arr, return true if it is possible to reorder arr such that arr[2 i + 1] = 2 arr[2 i] for every 0 <= i < len(arr) / 2, or false otherwise. Example 1: Input: arr = [3,1,3,6] Output: false Example 2: Input: arr = [2,1,2,6] Output: false Example 3: Input: arr = [4,-2,2,-4] Output: true Explanation: We can take two groups, [-2,-4] and [2,4] to form [-2,-4,2,4] or [2,4,-2,-4]. Constraints: 2 <= arr.length <= 3 104 arr.length is even. -105 <= arr[i] <= 105

Explanation

Here's an efficient solution to determine if an array can be reordered according to the given condition:

• Frequency Counting: Use a dictionary (or Counter) to store the frequency of each number in the input array.
• Iterative Matching: Iterate through the sorted keys of the frequency map. For each number x, check if 2 * x exists in the map and if their frequencies allow for pairing. Handle zero separately.

• Zero Handling: Special handling for zero because 2*0 = 0, if the count of zeros are odd, then return False.

• Runtime Complexity: O(n log n), where n is the length of the array, due to sorting the keys of the frequency map.

• Storage Complexity: O(n) in the worst case, to store the frequency map.

Code

    from collections import Counter

def canReorderDoubled(arr):
    """
    Determines if an array can be reordered such that arr[2 * i + 1] = 2 * arr[2 * i]
    for every 0 <= i < len(arr) / 2.

    Args:
        arr: An integer array of even length.

    Returns:
        True if it is possible to reorder arr as described, False otherwise.
    """
    counts = Counter(arr)
    for x in sorted(counts.keys(), key=abs):
        if counts[x] == 0:
            continue
        if counts[2 * x] < counts[x]:
            return False
        counts[2 * x] -= counts[x]
        counts[x] = 0
    return True