Tallest Billboard

public class Solution {
    public int maxHeightBalanced(int[] rods) {
        // map to store the difference between taller and shorter and the height of the taller
        Map<Integer, Integer> tally = new HashMap<>();
        tally.put(0, 0);
            
        for (int rod : rods) {
            // Temporary storage not to affect current iteration 
            Map<Integer, Integer> updatedTally = new HashMap<>(tally);
                
            for (Map.Entry<Integer, Integer> node : tally.entrySet()) {
                int difference = node.getKey();
                int tallerHeight = node.getValue();
                int shorterHeight = tallerHeight - difference;
                    
                // Assign the rod to the taller stack, increasing the size of the taller
                int updatedTallerHeight = updatedTally.getOrDefault(difference + rod, 0);
                updatedTally.put(difference + rod, Math.max(updatedTallerHeight, tallerHeight + rod));
                    
                // Adding rod to the shorter stack
                int newDifference = Math.abs(shorterHeight + rod - tallerHeight);
                int updatedTallestHeight = Math.max(shorterHeight + rod, tallerHeight);
                updatedTally.put(newDifference, Math.max(updatedTallestHeight, updatedTally.getOrDefault(newDifference, 0)));
            }
                
            tally = updatedTally;
        }
            
        // Return tallest height when the diference between taller and shorter is 0
        return tally.getOrDefault(0, 0);
    }
}

The provided solution in Java addresses the problem of constructing the tallest billboard using a set of rods, where the height of two sides of the billboard must be as equal as possible. This is a dynamic programming challenge that involves balancing the heights using the available rods represented as an integer array.

• Start by initializing a map named tally to store the difference between the heights of the two sides of the billboard (as keys) and the maximum height of the taller side when this difference is achieved (as values). Initially, this map contains the key-value pair (0, 0) denoting no difference and zero height.

• Iterate over each rod provided in the array. During each iteration, create a temporary map updatedTally to hold interim results. This avoids altering the entries of tally while still looping through it.

• For each entry in tally, calculate the potential new differences and heights if the current rod were to be added to either side:

  • If adding the rod to the taller side, update updatedTally to reflect this change. Determine if this new stacked height is maximal for the updated difference after adding the rod.
  • If adding the rod to the shorter side, calculate the new potential difference and update heights accordingly. Choose the greatest possible height for each difference value.

• After processing all rods, finalize the tally map with the values of updatedTally.

• The solution to the problem, i.e., the maximum height of a balanced billboard where the heights of the two sides are equal (difference of zero), is stored in tally under the key 0. Retrieve and return this value.

This method efficiently keeps track of all possible height imbalances and their associated maximum taller side height, dynamically adjusting as each rod is considered. The mapping strategy ensures that the complexity remains manageable, and no possibilities are missed.

class Solution:
    def maxEqualHeight(self, rods: List[int]) -> int:
        # Store the state
        state = {0:0}
            
        for rod in rods:
            # Avoid modifying the state during iteration
            updated_state = state.copy()
            for difference, max_height in state.items():
                smaller_height = max_height - difference
                    
                # Increase the height of the taller pile by the length of the current rod
                updated_state[difference + rod] = max(updated_state.get(difference + rod, 0), max_height + rod)
                    
                # Determine the new difference and maximum height if rod added to shorter pile
                new_difference = abs(smaller_height + rod - max_height)
                new_max_height = max(smaller_height + rod, max_height)
                updated_state[new_difference] = max(updated_state.get(new_difference, 0), new_max_height)
                    
            state = updated_state
                
        # Get the maximum equal height which is represented by the 0 difference
        return state.get(0, 0)

The provided code in Python3 solves the problem of determining the maximum height of two equal-height piles that can be formed by dividing a given set of rods. It utilizes dynamic programming to keep track of possible differences between the two piles and their associated maximum heights. Here’s a concise breakdown of the approach:

• Initialize a dictionary (state) to hold the difference between pile heights as keys, and their corresponding maximum possible taller pile height as values. Start with a base case of a zero difference and a height of zero.

• Iterate over each rod. For each rod, make a copy of the current state to prevent modifying the state during iteration.

• For each entry in the state (difference between piles and the maximum height of the taller pile):

  • Calculate the height of the shorter pile.
  • Attempt to add the current rod to the taller pile and update the new difference and maximum height if this results in a taller possible height.
  • Consider adding the rod to the shorter pile, recalculating the difference, and updating the maximum possible height.

• Continuously update the state dictionary with these new calculated values for each rod.

• At the end of iteration, maximum equal height of the two piles is the value in the state dictionary corresponding to a difference of zero.

The problem effectively seeks an arrangement where two subsets of the original rod lengths have the most closely matching sums, and this implementation dynamically calculates these optimal subsets by considering each rod incrementally.