
Problem Statement
When constructing a billboard, the goal is to achieve the maximum height for aesthetic and practical reasons. To support the billboard, two steel supports are required on either side, and critically, these supports must be of equal height for stability. You are given a variety of rods that can be welded together to create these supports.
The challenge lies in selecting and welding together subsets of these rods such that the two supports on either side of the billboard are of exactly equal height to maximize the billboard's height. The function you will create should return the maximum height possible for the billboard by determining the optimal combination of rods to form two supports of equal height. If no such combination is possible (meaning that no two subsets of rods can form supports of the same height), the function returns 0
.
Examples
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
Approach and Intuition
To solve this problem, the essential task is to identify two subsets of rods from the given collection that have an equal sum. This is a variation of the "Partition Problem", which is a well-known problem in computer science, specifically in the area of combinatorics and optimization.
The approach to this problem can be broken down as follows:
Initial Checks: Evaluate if it's feasible to split the total sum of all rods into two equal parts. If the sum of all rod lengths is odd, then it's immediately clear that forming two equal subsets is impossible, hence return
0
.Subset Sum Calculation: Utilize a dynamic programming approach to determine possible subsets and their sums. This involves creating a data structure, typically a set or array, that maintains these sums as states are transitioned through the addition of each rod.
Dynamic State Transition: For each rod that is considered, update the subsets sums by considering:
- Adding the rod to the existing subsets to form new subset possibilities.
- Ignoring the rod to maintain existing subsets.
Optimality Check: After evaluating all rods, the maximum value in the subset sum that equals half the total sum of all rods (if it exists) will be the maximum possible height.
Result Evaluation: The result in terms of the billboard height would be either twice this optimal subset sum (as each support would be of this maximum subset length) or
0
if no such subset can be found.
Given that:
- The constraints limit the number of rods to a maximum of 20 and their individual lengths to a maximum of 1000,
- Hence, a dynamic programming solution is feasible in terms of computational complexity.
The complexity mainly arises from the fact that every rod can potentially alter each subset sum previously computed, leading to exponential growth based on the number of rods. Thus, efficient management of data structures in the dynamic programming approach is crucial for optimizing performance.
Solutions
- Java
- Python
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 oftally
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.
- If adding the rod to the taller side, update
After processing all rods, finalize the
tally
map with the values ofupdatedTally
.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.
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