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// 1000. Minimum Cost to Merge Stones
#include "leetcode.h"
/*
* there are 'n' piles of 'stones' arrange in a row. the i'th pile has
* 'stones[i]' stones. a move consists of merging exactly 'k' consecutive piles
* into one pile and the cost of this move is equal to the total number of
* stones in these 'k' piles. return the minimum cost to merge all piles of
* stones into one pile. if it is impossible, return -1
*/
#define MAX_CNT 20
int mergeStones(int *stones, int stonesSize, int k) {
if ((stonesSize - 1) % (k - 1))
return -1;
int sum = 0, dp[MAX_CNT][MAX_CNT] = {0}, sum_arr[MAX_CNT + 1] = {0};
for (int i = 0; i < stonesSize; i++)
sum = sum_arr[i] = sum + stones[i];
for (int i = 0; i < stonesSize; i++) {
int j = i + k - 1;
if (j < stonesSize) {
dp[i][j] = sum_arr[j];
if (i)
dp[i][j] -= sum_arr[i - 1];
}
}
for (int i = 0; i < stonesSize - k; i++)
for (int j = 0, n = k + i; n < stonesSize; j++, n++) {
dp[j][n] = INT_MAX;
for (int mid = j; mid < n; mid += (k - 1)) {
int res = dp[j][mid] + dp[mid + 1][n];
dp[j][n] = res < dp[j][n] ? res : dp[j][n];
}
if (!((n - j) % (k - 1))) {
dp[j][n] += sum_arr[j];
if (j)
dp[j][n] -= sum_arr[j - 1];
}
}
return dp[0][stonesSize - 1];
}
int main() {
int s1[] = {3, 2, 4, 1}, s2[] = {3, 2, 4, 1}, s3[] = {3, 5, 1, 2, 6};
printf("%d\n", mergeStones(s1, ARRAY_SIZE(s1), 2)); // expect: 20
printf("%d\n", mergeStones(s2, ARRAY_SIZE(s2), 3)); // expect: -1
printf("%d\n", mergeStones(s3, ARRAY_SIZE(s3), 3)); // expect: 25
}
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