diff options
Diffstat (limited to 'max_num_point_cost.c')
-rw-r--r-- | max_num_point_cost.c | 46 |
1 files changed, 46 insertions, 0 deletions
diff --git a/max_num_point_cost.c b/max_num_point_cost.c new file mode 100644 index 0000000..8f1e97a --- /dev/null +++ b/max_num_point_cost.c @@ -0,0 +1,46 @@ +// 1937. Maximum Number of Points with Cost +#include "leetcode.h" + +/* + * you are given an 'm * n' integer matrix 'points' (0 indexed). starting with 0 + * points you want to maximise the numbber of points you can get from the + * matrix. to gain points, you must pick one cell in each row. picking the cell + * at coordinates '(r, c)' will add 'points[r][c]' to your score. however, you + * will lose points if you pick a cell too far from the cell that you picked in + * the previous row. for every two adjacent rows 'r' and 'r + 1', picking cells + * at coordinates '(r, c_1)' and '(r + 1, c_2)' will subtract 'abs(c_1 - c_2)' + * from your score. return the maximum number of points you can achieve + */ + +long long maxPoints(int **points, int pointsSize, int *pointsColSize) { + long long *pre = (long long *)malloc(*pointsColSize * sizeof(int)); + long long *left = (long long *)calloc(*pointsColSize, sizeof(long long)); + long long *right = (long long *)calloc(*pointsColSize, sizeof(long long)); + long long *curr = (long long *)calloc(*pointsColSize, sizeof(long long)); + for (int i = 0; i < *pointsColSize; i++) + pre[i] = points[0][i]; + for (int i = 0; i < pointsSize - 1; ++i) { + left[0] = pre[0]; + right[*pointsColSize - 1] = pre[*pointsColSize - 1]; + for (int j = 1; j < *pointsColSize; ++j) + left[j] = fmax(left[j - 1] - 1, pre[j]); + for (int j = *pointsColSize - 2; j >= 0; --j) + right[j] = fmax(right[j + 1] - 1, pre[j]); + for (int j = 0; j < *pointsColSize; ++j) + curr[j] = points[i + 1][j] + fmax(left[j], right[j]); + pre = curr; + } + long long ans = 0; + for (int i = 0; i < *pointsColSize; i++) + ans = fmax(ans, pre[i]); + free(pre), free(left), free(right), free(curr); + return ans; +} + +int main() { + int p1[3][3] = {{1, 2, 3}, {1, 5, 1}, {3, 1, 1}}, + p2[3][2] = {{1, 5}, {2, 3}, {4, 2}}, pcs1[] = {3, 3, 3}, + pcs2[] = {2, 2, 2}; + printf("%lld\n", maxPoints((int **)p1, ARRAY_SIZE(p1), pcs1)); // expect: 9 + printf("%lld\n", maxPoints((int **)p2, ARRAY_SIZE(p2), pcs2)); // expect: 11 +} |