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This approach involves using dynamic programming to calculate the maximum possible height of two equal steel supports from given rods. By defining a state that represents the difference of two potential support heights, we can get the maximum height efficiently. The key is to keep track of possible height differences between the two supports as we iterate over each rod.
Time Complexity: O(n * sum), where n is the number of rods and sum is the total length of rods.
Space Complexity: O(sum), where sum is the total length of rods.
1import java.util.HashMap;
2
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The Java solution follows the same algorithmic principle as the Python one. We maintain a map dp to track the maximum achievable height for each difference in support heights. For each rod, we consider adding it to both hypothetical sets and update the map accordingly.
This approach involves using dynamic programming with a difference array to explore the possible configurations of rods. We maintain a difference map where the key is the difference in height between two groups, and the value is the maximum sum achievable for that difference.
The idea here is to update the difference map as we iterate through each rod, considering it can contribute to either group or be omitted.
Time Complexity: O(N * S)
Space Complexity: O(S)
where N is the number of rods and S is the sum of all rods.
1#include <vector>
2#include <unordered_map>
3#include <algorithm>
4
5using namespace std;
6
7int tallestBillboard(vector<int>& rods) {
8 unordered_map<int, int> dp = {{0, 0}};
9 for (int rod : rods) {
10 unordered_map<int, int> new_dp = dp;
11 for (auto [d, y] : dp) {
12 new_dp[d + rod] = max(new_dp[d + rod], y);
13 new_dp[abs(d - rod)] = max(new_dp[abs(d - rod)], y + min(d, rod));
14 }
15 dp = new_dp;
16 }
17 return dp[0];
18}The C++ solution utilizes an unordered_map to maintain the difference states. With each rod, the map is updated by creating new states considering the current rod for each existing state.
This approach uses recursive backtracking along with memoization to explore all possible combinations of placing rods into two groups while aiming to minimize the difference between their sum. An auxiliary function is used to recursively try each rod in either of the sets or ignoring it, memoizing results to prevent redundant calculations.
Time Complexity: O(N * S)
Space Complexity: O(N * S) for the memoization cache.
1from functools import lru_cache
2
3def tallest_billboard(rods):
4 n = len(rods)
5
6 @lru_cache(None)
7 def dfs(index, diff):
8 if index == n:
9 return 0 if diff == 0 else float('-inf')
10 add_to_first = dfs(index + 1, diff + rods[index]) + rods[index]
11 add_to_second = dfs(index + 1, diff - rods[index])
12 skip = dfs(index + 1, diff)
13 return max(add_to_first, add_to_second, skip)
14
15 return dfs(0, 0)The recursive function dfs explores all possible assignments of rods into the two groups, with memoization applied using lru_cache to store intermediate result computations. This function seeks to balance the groups while maximizing the sum.