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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.
1#include <stdio.h>
2#include <stdlib.h>
3#include <limits.h>
4
5int minCost(int n, int* cuts, int cutsSize) {
6 qsort(cuts, cutsSize, sizeof(int), (int(*)(const void*, const void*))strcmp);
7 int* newCuts = (int*)malloc((cutsSize + 2) * sizeof(int));
8 newCuts[0] = 0; newCuts[cutsSize + 1] = n;
9 for (int i = 0; i < cutsSize; ++i) newCuts[i + 1] = cuts[i];
10 int** dp = (int**)malloc((cutsSize + 2) * sizeof(int*));
11 for (int i = 0; i < cutsSize + 2; ++i) {
12 dp[i] = (int*)malloc((cutsSize + 2) * sizeof(int));
13 for (int j = 0; j < cutsSize + 2; ++j) dp[i][j] = 0;
14 }
15 for (int len = 2; len <= cutsSize + 1; ++len) {
16 for (int i = 0; i <= cutsSize + 1 - len; ++i) {
17 int j = i + len;
18 dp[i][j] = INT_MAX;
19 for (int k = i + 1; k < j; ++k) {
20 dp[i][j] = fmin(dp[i][j], newCuts[j] - newCuts[i] + dp[i][k] + dp[k][j]);
21 }
22 }
23 }
24 int result = dp[0][cutsSize + 1];
25 for (int i = 0; i < cutsSize + 2; ++i) free(dp[i]);
26 free(dp);
27 free(newCuts);
28 return result;
29}
30
31int main() {
32 int cuts[] = {1, 3, 4, 5};
33 int n = 7;
34 printf("Minimum cost: %d\n", minCost(n, cuts, 4));
35 return 0;
36}This C solution sorts the cuts and inserts them along with the stick boundaries into a new array 'newCuts'. The dynamic programming table 'dp' is used to record minimum costs for every subproblem defined by start and end positions in the newCuts array. The cost of each subproblem is solved by iterating over possible cut positions within that range and selecting the smallest achievable cost.
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.