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+// 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
+}