// 3494. Find the Minimum Amount of Time to Brew Potions #include "leetcode.h" /* * you are given two integer arrays 'skill' and 'mana' of length 'n' and 'm' * respectively. in a laboratory, 'n' wizards must brew 'm' potions in order. * each potion has a mana capacity 'mana[j]' and must pass through all the * wizards sequentially to be brewed properly. the time taken by the i'th wizard * on the j'th potion is 'time_ij = skill[i] * mana[j]'. since the brewing * process is delicate, a potion must be passed to the next wizard immediately * after the current wizard completes their work. this means the timing must be * synchronised so that each wizard begins working on a potion exactly when it * arrives. return the minimum amount of time required for the potions to be * brewed properly. */ long long minTime(int *skill, int skillSize, int *mana, int manaSize) { long long *pre = (long long *)calloc(skillSize + 1, sizeof(long long)); for (int i = 1; i < skillSize; i++) pre[i] = pre[i - 1] + skill[i]; long long sum = (long long)skill[0] * mana[0]; for (int j = 1; j < manaSize; j++) { long long max = (long long)skill[0] * mana[j]; for (int i = 1; i < skillSize; i++) { long long diff = pre[i] * mana[j - 1] - pre[i - 1] * mana[j]; if (diff > max) max = diff; } sum += max; } int ans = pre[skillSize - 1] * mana[manaSize - 1]; free(pre); return sum + ans; } int main() { int s1[] = {1, 5, 2, 4}, m1[] = {5, 1, 4, 2}; int s2[] = {1, 1, 1}, m2[] = {1, 1, 1}; int s3[] = {1, 2, 3, 4}, m3[] = {1, 2}; printf("%lld\n", minTime(s1, ARRAY_SIZE(s1), m1, ARRAY_SIZE(m1))); // expect: 110 printf("%lld\n", minTime(s2, ARRAY_SIZE(s2), m2, ARRAY_SIZE(m2))); // expect: // 5 printf("%lld\n", minTime(s3, ARRAY_SIZE(s3), m3, ARRAY_SIZE(m3))); // expect: 21 }