Sponsored
Sponsored
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.
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 public int TallestBillboard(int[] rods) {
6 Dictionary<int, int> dp = new Dictionary<int, int>() {{ 0, 0 }};
7 foreach (var rod in rods) {
8 var newDp = new Dictionary<int, int>(dp);
9 foreach (var kvp in dp) {
10 int d = kvp.Key;
11 int y = kvp.Value;
12 if (!newDp.ContainsKey(d + rod)) newDp[d + rod] = 0;
13 newDp[d + rod] = Math.Max(newDp[d + rod], y);
14 if (!newDp.ContainsKey(Math.Abs(d - rod))) newDp[Math.Abs(d - rod)] = 0;
15 newDp[Math.Abs(d - rod)] = Math.Max(newDp[Math.Abs(d - rod)], y + Math.Min(d, rod));
16 }
17 dp = newDp;
18 }
19 return dp.ContainsKey(0) ? dp[0] : 0;
20 }
21}The C# solution uses a Dictionary to maintain the dynamic programming states as key-value pairs, much akin to implementations in Java and C++.
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.
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 private Dictionary<string, int> memo = new Dictionary<string, int>();
6
7 public int TallestBillboard(int[] rods) {
8 return Dfs(rods, 0, 0);
9 }
10
11 private int Dfs(int[] rods, int index, int diff) {
12 if (index == rods.Length) return diff == 0 ? 0 : int.MinValue;
13
14 string key = index + "," + diff;
15 if (memo.ContainsKey(key)) return memo[key];
16
17 int addToFirst = Dfs(rods, index + 1, diff + rods[index]) + rods[index];
18 int addToSecond = Dfs(rods, index + 1, diff - rods[index]);
19 int skip = Dfs(rods, index + 1, diff);
20
21 int result = Math.Max(Math.Max(addToFirst, addToSecond), skip);
22 memo[key] = result;
23 return result;
24 }
25}This C# method employs recursion with a Dictionary for memoization, handling each step and decision recursively through depth-first search to balance rod groups efficiently.