Introduction

In this blog post, we will explore how to solve the LeetCode problem "3309" 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 of integers nums of size 3. Return the maximum possible number whose binary representation can be formed by concatenating the binary representation of all elements in nums in some order. Note that the binary representation of any number does not contain leading zeros. Example 1: Input: nums = [1,2,3] Output: 30 Explanation: Concatenate the numbers in the order [3, 1, 2] to get the result "11110", which is the binary representation of 30. Example 2: Input: nums = [2,8,16] Output: 1296 Explanation: Concatenate the numbers in the order [2, 8, 16] to get the result "10100010000", which is the binary representation of 1296. Constraints: nums.length == 3 1 <= nums[i] <= 127

Explanation

Here's the solution:

• High-Level Approach: The core idea is to try all possible permutations of the input numbers. For each permutation, we concatenate the binary representations of the numbers in that order and convert the resulting binary string to an integer. Finally, we return the maximum integer obtained across all permutations. Since the array size is fixed at 3, we can simply generate all 3! = 6 permutations directly without recursion or external permutation functions.

• Complexity: O(1) time complexity & O(1) space complexity. This is because the number of operations and the amount of memory used are bounded by a constant since the input array size is fixed at 3.

Code

    def max_concatenated_binary(nums):
    """
    Finds the maximum possible number whose binary representation can be formed by
    concatenating the binary representation of all elements in nums in some order.

    Args:
        nums: A list of integers.

    Returns:
        The maximum possible number.
    """

    def to_binary(n):
        return bin(n)[2:]

    a, b, c = nums
    perms = [
        [a, b, c],
        [a, c, b],
        [b, a, c],
        [b, c, a],
        [c, a, b],
        [c, b, a],
    ]

    max_val = 0
    for perm in perms:
        binary_string = "".join(to_binary(x) for x in perm)
        max_val = max(max_val, int(binary_string, 2))

    return max_val