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This method leverages dynamic programming to find the minimum cost by solving subproblems. We start by sorting the cuts to facilitate an easier partitioning strategy. Then, define a dp table where dp[i][j] represents the minimum cost to perform all the cuts between positions i and j.
Use the recursive relation to compute the minimum cost for each partition (i, j) by considering every possible cut position in that interval. The base case is that if there are no cuts between two positions, the cost is zero.
Time Complexity: O(m^3) where m is the number of cuts, due to the nested loops solving subproblems.
Space Complexity: O(m^2) for the dp table storage, where m is the number of cuts.
1import java.util.*;
2
3class Solution {
4 public int minCost(int n, int[] cuts) {
5 Arrays.sort(cuts);
6 int m = cuts.length;
7 int[] newCuts = new int[m + 2];
8 System.arraycopy(cuts, 0, newCuts, 1, m);
9 newCuts[0] = 0;
10 newCuts[m + 1] = n;
11 int[][] dp = new int[m + 2][m + 2];
12 for (int len = 2; len <= m + 1; ++len) {
13 for (int i = 0; i <= m + 1 - len; ++i) {
14 int j = i + len;
15 dp[i][j] = Integer.MAX_VALUE;
16 for (int k = i + 1; k < j; ++k) {
17 dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k][j] + newCuts[j] - newCuts[i]);
18 }
19 }
20 }
21 return dp[0][m + 1];
22 }
23
24 public static void main(String[] args) {
25 Solution sol = new Solution();
26 int[] cuts = {1, 3, 4, 5};
27 int n = 7;
28 System.out.println("Minimum cost: " + sol.minCost(n, cuts));
29 }
30}This Java solution closely mirrors the logic of the C and C++ examples. It uses the Arrays utility to sort the cuts and define boundaries before filling in the dp table. This solution optimally calculates the minimum cost of cutting with nested loops on the 'newCuts' set.
This approach uses recursive function calls along with a memoization hash table to store the results of subproblem calculations. This reduces repeated calculations and enhances efficiency. Order the cuts, then define a recursive function that selects the appropriate cut division and employs the same function to solve newly formed partitions recursively.
Time Complexity: O(m^2) with memoization staving off excessive recomputation.
Space Complexity: O(m^2) result-caching leading to rational storage expansion.
1import java.util.*;
2
Java's HashMap is adopted for caching operations within this solution, leveraging recursion through a cost function to deliver minimized cutting outcomes by memorizing intermediate results and returning final stored outcomes for given input sequences.