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author | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
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committer | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /max_score_perform_op.c | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
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Diffstat (limited to 'max_score_perform_op.c')
-rw-r--r-- | max_score_perform_op.c | 39 |
1 files changed, 39 insertions, 0 deletions
diff --git a/max_score_perform_op.c b/max_score_perform_op.c new file mode 100644 index 0000000..745127d --- /dev/null +++ b/max_score_perform_op.c @@ -0,0 +1,39 @@ +// 1770. Maximum Score from Performing Multiplication Operations +#include "leetcode.h" + +/* + * given two 0 indexed integer arrays 'nums' and multipliers of size n and m + * respectively, where n >= m. you begin with a score of 0. you want to perform + * exactly m operations. on the i'th operation, you will choose one integer x + * from either the start or the end of the array 'nums'. add 'multipliers[i] * + * x' to your score. note that 'multipliers[0]' correspons to the first + * operation, 'multipliers[1]' to the second operation and so on. return the + * maximum score after performing m operations. + */ + +int maximumScore(int *nums, int numsSize, int *multipliers, + int multipliersSize) { + int dp_size = fmin(numsSize, multipliersSize); + int *row = &nums[numsSize - dp_size], *col = nums; + int **dp = malloc(dp_size * sizeof(int *)); + for (int i = 0; i < dp_size; i++) { + dp[i] = malloc(dp_size * sizeof(int)); + dp[i][i] = fmax(multipliers[dp_size - 1] * row[i], + multipliers[dp_size - 1] * col[i]); + } + for (int i = 1; i < dp_size; i++) + for (int j = 0; j < dp_size - i; j++) + dp[j][i + j] = + fmax(col[j] * multipliers[dp_size - 1 - i] + dp[j + 1][i + j], + row[i + j] * multipliers[dp_size - 1 - i] + dp[j][i + j - 1]); + return dp[0][dp_size - 1]; +} + +int main() { + int n1[] = {1, 2, 3}, m1[] = {3, 2, 1}, n2[] = {-5, -3, -3, -2, 7, 1}, + m2[] = {-10, -5, 3, 4, 6}; + printf("%d\n", + maximumScore(n1, ARRAY_SIZE(n1), m1, ARRAY_SIZE(m1))); // expect: 14 + printf("%d\n", + maximumScore(n2, ARRAY_SIZE(n2), m2, ARRAY_SIZE(m2))); // expect: 102 +} |