// 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 }