Introduction

In this blog post, we will explore how to solve the LeetCode problem "956" 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 installing a billboard and want it to have the largest height. The billboard will have two steel supports, one on each side. Each steel support must be an equal height. You are given a collection of rods that can be welded together. For example, if you have rods of lengths 1, 2, and 3, you can weld them together to make a support of length 6. Return the largest possible height of your billboard installation. If you cannot support the billboard, return 0. Example 1: Input: rods = [1,2,3,6] Output: 6 Explanation: We have two disjoint subsets {1,2,3} and {6}, which have the same sum = 6. Example 2: Input: rods = [1,2,3,4,5,6] Output: 10 Explanation: We have two disjoint subsets {2,3,5} and {4,6}, which have the same sum = 10. Example 3: Input: rods = [1,2] Output: 0 Explanation: The billboard cannot be supported, so we return 0. Constraints: 1 <= rods.length <= 20 1 <= rods[i] <= 1000 sum(rods[i]) <= 5000

Explanation

Here's a breakdown of the approach and the Python code:

• Key Idea: The problem is equivalent to partitioning the rods into two subsets with equal sums (representing the heights of the supports). A difference variable is maintained. We iterate through the rods, and for each rod, we have three choices: add it to the first support, add it to the second support, or don't use it.
• Dynamic Programming: Use dynamic programming to store and reuse solutions to subproblems. The state represents the index of the rod currently being considered and the difference between the sums of the two supports.

• Optimization: Use memoization to avoid recomputing the same subproblems.

• Complexity: O(n sum), where n is the number of rods and sum is the sum of all rod lengths. The space complexity is also O(n sum).

Code

    def tallestBillboard(rods):
    """
    Finds the largest possible height of a billboard installation with two equal-height supports.

    Args:
        rods: A list of integers representing the lengths of the rods.

    Returns:
        The largest possible height of the billboard, or 0 if it cannot be supported.
    """
    total_sum = sum(rods)
    memo = {}

    def dp(index, diff):
        """
        Recursive helper function to calculate the maximum height.

        Args:
            index: The index of the current rod being considered.
            diff: The difference between the sums of the two supports.

        Returns:
            The maximum possible height.
        """
        if index == len(rods):
            return 0 if diff == 0 else float('-inf')

        if (index, diff) in memo:
            return memo[(index, diff)]

        # Option 1: Don't use the current rod
        option1 = dp(index + 1, diff)

        # Option 2: Add the current rod to the first support
        option2 = dp(index + 1, diff + rods[index])

        # Option 3: Add the current rod to the second support
        option3 = dp(index + 1, diff - rods[index])

        memo[(index, diff)] = max(option1, option2, option3)
        return memo[(index, diff)]

    result = dp(0, 0)
    return result if result > 0 else 0