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Diffstat (limited to 'mini_maxi_products_distrib.c')
-rw-r--r-- | mini_maxi_products_distrib.c | 45 |
1 files changed, 45 insertions, 0 deletions
diff --git a/mini_maxi_products_distrib.c b/mini_maxi_products_distrib.c new file mode 100644 index 0000000..7ea4016 --- /dev/null +++ b/mini_maxi_products_distrib.c @@ -0,0 +1,45 @@ +// 2064. Minimized Maximum of Products Distributed to Any Store +#include "leetcode.h" + +/* + * you are given an integer 'n' indicating that there are 'n' specialty retail + * stores. there are 'm' product types of varying amounts, which are given as a + * 0-indexed integer array 'quantities' owhere 'quantities[i]' represents the + * number of products of the i'th product type. you need to distribute all the + * products to the retail stores following these rules: a store can only be + * given at most one product type but can be given any amount of it, and after + * distribution, each store will have been given some number products. let 'x' + * represent the maximum number of products given to any store. you want 'x' to + * be as small as possible. return the minimum possible 'x'. + */ + +int minimizedMaximum(int n, int *quantities, int quantitiesSize) { + long long sum = 0; + int right = 0; + for (int i = 0; i < quantitiesSize; ++i) { + sum += quantities[i]; + if (right < quantities[i]) + right = quantities[i]; + } + if (sum < n) + return 1; + int left = (int)(sum / n), mid = 0, groups = 0; + while (left <= right) { + mid = left + (right - left) / 2; + groups = 0; + for (int i = 0; i < quantitiesSize; i++) + groups += (quantities[i] + mid - 1) / mid; + if (groups <= n) + right = mid - 1; + else + left = mid + 1; + } + return left; +} + +int main() { + int q1[] = {11, 6}, q2[] = {15, 10, 10}, q3[] = {100000}; + printf("%d\n", minimizedMaximum(6, q1, ARRAY_SIZE(q1))); // expect: 3 + printf("%d\n", minimizedMaximum(7, q2, ARRAY_SIZE(q2))); // expect: 5 + printf("%d\n", minimizedMaximum(1, q3, ARRAY_SIZE(q3))); // expect: 100000 +} |