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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.
1from collections import defaultdict
2
3Solve with full IDE support and test cases
We use a dictionary dp to store the maximum possible height of the smaller support for each possible height difference. For each rod, we have two choices: add the rod to one support or to the other. We update the dp table accordingly by creating a copy and updating from it to avoid overwriting during iteration.
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.
1def tallest_billboard(rods):
2 dp = {0: 0}
3 for rod in rods:
4 new_dp = dp.copy()
5 for d, y in dp.items():
6 new_dp[d + rod] = max(new_dp.get(d + rod, 0), y)
7 new_dp[abs(d - rod)] = max(new_dp.get(abs(d - rod), 0), y + min(d, rod))
8 dp = new_dp
9 return dp[0]A dictionary dp is maintained where each key denotes a possible difference between the height of two supports, and the value denotes the maximum height obtained for this difference. The new height differences are updated for each rod considered, creating new states for the DP table.
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.
1#include <vector>
2#include <unordered_map>
3#include <algorithm>
4
5using namespace std;
6
7class Solution {
8 int dfs(const vector<int>& rods, int index, int diff, unordered_map<string, int>& memo) {
9 if (index == rods.size())
10 return diff == 0 ? 0 : INT_MIN;
11 string key = to_string(index) + "," + to_string(diff);
12 if (memo.find(key) != memo.end())
13 return memo[key];
14
15 int add_to_first = dfs(rods, index + 1, diff + rods[index], memo) + rods[index];
16 int add_to_second = dfs(rods, index + 1, diff - rods[index], memo);
17 int skip = dfs(rods, index + 1, diff, memo);
18
19 return memo[key] = max({add_to_first, add_to_second, skip});
20 }
21
22public:
23 int tallestBillboard(vector<int>& rods) {
24 unordered_map<string, int> memo;
25 return dfs(rods, 0, 0, memo);
26 }
27};Utilizing a recursive function dfs wrapped with memoization via a map, this C++ implementation mirrors the recursive approach from Python. It employs depth-first search to manage rod groupings distinctly.