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// 1547. Minimum Cost to Cut a Stick
#include "leetcode.h"
/*
* given a wooden stick of length 'n' units. the stick is labelled from 0 to
* 'n'. for example, a stick of length 6 is labelled as follows (see image).
* given an integer array 'cuts' where 'cuts[i]' denotes a position you should
* perform a cut at. you should perform the cuts in order, you can change the
* order of the cuts as you wish. the cost of one cut is the length of the stick
* to be cut, the total cost is the sum of costs of all cuts. when you cut a
* stick, it will be split into two smaller sticks ie. the sum of their lengths
* is the length of the stick before the cut. refer to first example for better
* explanation (see image). return minimum cost of the cuts
*/
class Solution {
int dp[102][102] = {};
int dfs(vector<int> &cuts, int i, int j) {
if (j - i <= 1)
return 0;
if (!dp[i][j]) {
dp[i][j] = INT_MAX;
for (auto k = i + 1; k < j; ++k)
dp[i][j] = min(dp[i][j],
cuts[j] - cuts[i] + dfs(cuts, i, k) + dfs(cuts, k, j));
}
return dp[i][j];
}
public:
int minCost(int n, vector<int> &cuts) {
cuts.push_back(0);
cuts.push_back(n);
sort(begin(cuts), end(cuts));
return dfs(cuts, 0, cuts.size() - 1);
}
};
int main() {
Solution obj;
vector<int> c1 = {1, 3, 4, 5};
vector<int> c2 = {5, 6, 1, 4, 2};
printf("%d\n", obj.minCost(7, c1)); // expect: 16
printf("%d\n", obj.minCost(9, c2)); // expect: 22
}
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