diff options
Diffstat (limited to 'min_cost_merge_stones.c')
-rw-r--r-- | min_cost_merge_stones.c | 49 |
1 files changed, 49 insertions, 0 deletions
diff --git a/min_cost_merge_stones.c b/min_cost_merge_stones.c new file mode 100644 index 0000000..000aa6f --- /dev/null +++ b/min_cost_merge_stones.c @@ -0,0 +1,49 @@ +// 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 +} |